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Unformatted text preview: Sketch Solutions for Exercises in the Main Text of A First Look at Vectors T. W. K¨rner o 1 2 Introduction Here are what I believe to be sketch solutions to the bulk of exercises in the main text the book (i.e. those not in the “Further Exercises”). I have written in haste in the hope that others will help me correct at leisure. I am sure that they are stuffed with errors ranging from the TEXtual through to the arithmetical and not excluding serious mathematical mistakes. I would appreciate the opportunity to correct at least some of these problems. Please tell me of any errors, unbridgeable gaps, misnumberings etc. I welcome suggestions for additions. ALL COMMENTS GRATEFULLY RECEIVED. A If you can, please use L TEX 2ε or its relatives for mathematics. If not, please use plain text. My e-mail is twk@dpmms.cam.ac.uk. You may safely assume that I am both lazy and stupid so that a message saying ‘Presumably you have already realised the mistake in Exercise Z ’ is less useful than one which says ‘I think you have made a mistake in Exercise Z because you have have assumed that the sum is necessarily larger than the integral. One way round this problem is to assume that f is decreasing.’ When I was young, I used to be surprised when the answer in the back of the book was wrong. I could not believe that the wise and gifted people who wrote textbooks could possibly make mistakes. I am no longer surprised. It may be easiest to navigate this document by using the table of contents which follow on the next few pages. To avoid disappointment, observe that those exercises marked ⋆ have no solution given. 3 Contents Introduction Exercise 1.1.2 Exercise 1.2.1 Exercise 1.2.2 Exercise 1.2.3 Exercise 1.2.5 Exercise 1.2.6 Exercise 1.2.7 Exercise 1.2.8 Exercise 1.2.9⋆ Exercise 1.2.10 Exercise 1.3.5 Exercise 1.3.9 Exercise 1.3.10 Exercise 1.4.3 Exercise 2.1.3 Exercise 2.1.4 Exercise 2.1.6⋆ Exercise 2.1.7⋆ Exercise 2.1.9 Exercise 2.2.4⋆ Exercise 2.2.5⋆ Exercise 2.2.8 Exercise 2.2.10 Exercise 2.3.3 Exercise 2.3.4⋆ Exercise 2.3.11 Exercise 2.3.12 Exercise 2.3.13 Exercise 2.3.14 Exercise 2.3.16 Exercise 2.3.17 Exercise 2.3.18 Exercise 2.4.1 Exercise 2.4.2 Exercise 2.4.4⋆ Exercise 2.4.6 Exercise 2.4.9 Exercise 2.4.10 Exercise 2.4.11 Exercise 2.4.12 Exercise 3.2.1 Exercise 3.3.4 2 13 14 15 16 17 18 19 20 20 21 22 23 25 26 27 28 28 28 29 29 29 30 31 32 32 33 34 35 36 37 38 39 40 41 41 42 43 44 45 46 47 48 4 Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise 3.3.6 3.3.9 3.3.10 3.3.12 3.3.13 3.3.14 3.4.5 3.4.7 3.5.1 3.5.2 4.1.1 4.1.2 4.1.3 4.2.1 4.2.2 4.2.3⋆ 4.3.2 4.3.3 4.3.10 4.3.13 4.3.14 4.3.15 4.3.16⋆ 4.4.1 4.4.4 4.4.6 4.4.7 4.4.8 4.4.9 4.5.1 4.5.3 4.5.3 4.5.5 4.5.6 4.5.7 4.5.8 4.5.10 4.5.13 4.5.14 4.5.15 4.5.16 4.5.17 5.1.1 5.1.2⋆ 5.2.7 49 50 52 53 54 55 56 57 58 59 60 61 62 63 64 64 65 66 67 68 69 70 70 71 72 73 74 75 76 78 79 80 81 82 83 84 85 86 87 88 89 90 91 91 92 5 Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise 5.2.8 5.2.11 5.3.2 5.3.3 5.3.10 5.3.11 5.3.14 5.3.16 5.4.3 5.4.11 5.4.12 5.4.13 5.4.14 5.5.3 5.5.5 5.5.6 5.5.12⋆ 5.5.13 5.5.14 5.5.16 5.5.17 5.5.18 5.5.19 5.6.2 5.6.3 5.6.4 5.6.5 5.6.6 6.1.2 6.1.3 6.1.5 6.1.7 6.1.8 6.1.9 6.1.12 6.2.3 6.2.4 6.2.9 6.2.11 6.2.14 6.2.15 6.2.16 6.3.2 6.4.2 6.4.5 93 94 95 96 97 98 99 100 101 102 104 105 106 107 108 109 109 110 111 113 114 115 116 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 6 Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise 6.4.7 6.4.8 6.4.9 6.5.5 6.5.2 6.6.2 6.6.7 6.6.8 6.6.9 6.6.10 6.6.11 6.6.12 6.6.13 6.7.1 6.7.2 6.7.3 6.7.6 6.7.7 6.7.9 6.7.10 6.7.11 6.7.12 7.1.4 7.1.6 7.1.8 7.1.9 7.1.10 7.2.2 7.2.6 7.2.10 7.2.13 7.3.2 7.3.4 7.3.6 7.3.7 7.4.3 7.4.6 7.4.9 7.5.1 7.5.2 7.5.3 7.5.4 7.5.5 7.5.6 7.5.7 140 141 142 143 144 145 146 147 148 149 150 151 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 185 186 7 Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise 8.1.7 8.2.2 8.2.6 8.2.7 8.2.9 8.2.10 8.3.2 8.3.3⋆ 8.3.4 8.3.5 8.4.2 8.4.3 8.4.4 8.4.6 8.4.7 8.4.8 8.4.9 8.4.10 8.4.11 8.4.12 8.4.13 8.4.14 8.4.15 8.4.16 8.4.17 8.4.18 8.4.19 8.4.20 9.1.2 9.1.4 9.2.1 9.3.1 9.3.5 9.3.6 9.3.8 9.4.3 9.4.5 9.4.6 9.4.7 9.4.9 9.4.10 9.4.11 9.4.12 9.4.13 9.4.14 188 189 190 191 192 193 194 194 195 196 197 198 199 200 201 202 203 204 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 225 226 227 228 229 230 231 232 233 8 Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise 9.4.15 9.4.16 9.4.17 9.4.19 9.4.21 9.4.22 10.1.4 10.1.2 10.1.3 10.1.6 10.1.8 10.2.1 10.2.2 10.2.3 10.2.4 10.3.2 10.4.1 10.4.3 10.4.6 11.1.2 11.1.3 11.1.4 11.1.5 11.1.9 11.1.15 11.1.16 11.2.1 11.2.3 11.2.4 11.2.5 11.3.1 11.3.6 11.3.7 11.3.8 11.3.9 11.3.10 11.4.2 11.4.4 11.4.5 11.4.6 11.4.9 11.4.10 11.4.12 11.4.15 11.4.19 234 235 236 237 238 239 240 241 242 243 245 246 247 248 249 250 251 252 253 254 255 256 257 258 260 261 262 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 9 Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise 11.4.20 12.1.2 12.1.3 12.1.4 12.1.6 12.1.7 12.1.11 12.1.13 12.1.14 12.1.15 12.1.16 12.1.17 12.1.18 12.1.19 12.2.1 12.2.2 12.2.3 12.2.6 12.2.7 12.2.8 12.2.9 12.2.11 12.2.12 12.3.1 12.3.3 12.3.6 12.3.7 12.3.9 12.3.11 12.3.13 12.3.14 12.3.15 12.4.3 12.4.4 12.4.7 12.4.11 12.4.12 12.4.13 12.4.14 12.5.1 12.5.2 12.5.3 12.5.4 12.5.5 12.5.6 284 285 286 287 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 316 317 318 319 320 321 322 323 324 326 327 328 330 331 332 10 Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise 12.5.7 12.5.8 12.5.9 12.5.10 12.5.11 13.1.2 13.1.3 13.1.4 13.1.6 13.1.7 13.1.8 13.1.9 13.2.1 13.2.2 13.2.3 13.2.4 13.2.5 13.2.6 13.2.7 13.2.9 13.2.10 13.2.12 13.2.13 13.2.14 13.2.15 13.2.16 13.2.17 13.2.18 13.3.3 13.3.4 13.3.5 13.3.11 13.3.12 13.3.14 13.3.16 13.4.1 13.4.3 13.4.5⋆ 13.4.6 13.4.7 13.4.10 13.4.12 13.5.2 13.5.3 13.5.4 333 334 336 337 338 340 341 342 343 344 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 367 369 370 372 373 374 375 375 376 377 378 379 380 381 382 11 Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise 13.5.5 13.5.7 13.5.8 13.5.9 13.5.10⋆ 13.5.11 13.5.12 13.5.13 13.5.14 13.5.15 14.1.1 14.1.3 14.1.7 14.1.9 14.1.10 14.1.11 14.1.12 14.1.13 14.2.2 14.2.4 14.3.2 14.3.3 14.3.7 14.3.8 14.3.11 14.4.3 14.4.6 14.4.7 14.4.9 15.1.8 15.1.13 15.1.15 15.1.16 15.1.22 15.1.24 15.1.25 15.1.26 15.2.4 15.2.8 15.2.9 15.2.10 15.2.12 15.2.13 15.3.2 15.3.3 383 384 385 386 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 408 409 411 412 413 414 415 416 418 419 420 422 423 424 425 426 428 429 430 431 12 Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise 15.3.4 15.3.5 15.3.6 15.3.7 15.3.11 15.3.12 15.3.14 15.3.15 15.3.16 15.3.17 15.3.19 15.4.2 15.4.3 15.4.6 15.4.7 15.4.8 15.4.9 432 433 434 436 437 438 439 440 441 443 444 446 447 448 449 450 451 13 Exercise 1.1.2 Starting with x+y+z =1 x + 2y + 3z = 2 x + 4y + 9z = 6. We subtract the first equation from the second and from the third to get x+y+z =1 y + 2z = 1 3y + 8z = 5. We now solve the equations y + 2z = 1 3y + 8z = 5, for z by subtracting 3 times the first equation from the second to obtain y + 2z = 2 2z = 2. Thus z = 1 and, working backwards, y = 1 − 2z = −1, whence x = 1 − y − z = 1. 14 Exercise 1.2.1 STEP 1 If aij = 0 for all i and j , then our equations have the form [1 ≤ i ≤ m]. 0 = yi Our equations are inconsistent unless y1 = y2 = . . . = ym = 0. If y1 = y2 = . . . = ym = 0 the equations impose no constraints on x1 , x2 , . . . , xn which can take any value we want. STEP 2 If the condition of STEP 1 does not hold, we can arrange,by reordering the equations and the unknowns, if necessary, that a11 = 0. We now subtract ai1 /a11 times the first equation from the ith equation [2 ≤ i ≤ m] to obtain n [2 ≤ i ≤ m]. bij xj = zi ⋆⋆ j =2 where bij = a11 aij − ai1 a1j a11 yi − ai1 y1 and zi = . a11 a11 STEP 3 If the new set of equations ⋆⋆ has no solution, then our old set ⋆ has no solution. If our new set of equations ⋆⋆ has a solution xi = x′i for 2 ≤ i ≤ m, then our old set ⋆ has the solution 1 x1 = a11 xi = x′i (Continued in Exercise 1.2.2.) n y1 − a1j x′j j =2 [2 ≤ i ≤ n]. 15 Exercise 1.2.2 (Continued from Exercise 1.2.1.) This means that, if ⋆⋆ has exactly one solution, then ⋆ has exactly one solution and, if ⋆⋆ has infinitely many solutions, then ⋆ has infinitely many solutions. We have already remarked that, if ⋆⋆ has no solutions, then ⋆ has no solutions. 16 Exercise 1.2.3 (i)(a) If ai = 0, but yi = 0 for some i, then there can be no solution, since the equation 0 = yi cannot be satisfied. (b) If yr = 0 whenever ar = 0 and there exists an i such that ai = 0, then there are an infinity of solutions with xs = a−1 ys when as = 0 and s ys chosen freely otherwise. (c) If ai = 0 for all i, there is a unique solution xi = a−1 yi for all i. i (ii) (a) If aj = 0 for all j then either b = 0 and every choice of xj gives a solution (so there is an infinity of solutions or b = 0 and there is no solution. (b) If ak = 0 for some k , then, choosing xi freely for i = k and setting x k = a− 1 b − k gives an infinity of solutions. aj x j j =k 17 Exercise 1.2.5 (i) If a = 1, b = 2, c = d = 4 the first and second equations are incompatible, so there is no solution. (ii) If a = 1, b = 2, c = d = 4, then the third equation gives the same information as the first equation, so the system reduces to x+y =2 x + 2y = 4 Subtracting the first equation from the second we see that y = 2. Thus x = 0. By inspection this is a solution. (iii) If a = b = 2, c = d = 4, then the second and third equation give the same information as the first equation so the system reduces to x+y =1 with the infinite set of solutions given by choosing x arbitrarily and setting y = 1 − x. 18 Exercise 1.2.6 The two equations are incompatible if a = 1. There is then no solution. If a = 1, then subtracting the first equation from the second we get x+y+z =2 (a − 1)z = 2, so z = 2/(a − 1). Knowing the value of z the system reduces to 2 a−2 x+y =2− =2 a−1 a−1 so x may be chosen freely and then a−2 − x. y=2 a−1 There are an infinity of solutions. 19 Exercise 1.2.7 (i) Observe that, if n ≥ 4 n n3 ≥ n r =1 n2 ≥ r2 ≥ r =1 n (n/2)2 ≥ (n/4) × (n/2)2 = n3 /16. n≥r ≥n/2 (ii) Let f (x) = x2 . Then f is increasing so f (r) ≤ f (x) ≤ f (r + 1) for r ≤ x ≤ r + 1. Integrating we get f (r) = r r +1 r +1 r +1 f (r) dx ≤ f (x) dx ≤ r f (r + 1) dx = f (r + 1) r Thus, summing, n−1 r =1 In other words n n f (r) ≤ f (x) dx ≤ 1 f (r). r =2 n−1 n 2 r =1 so 3 r ≤ (n − 1)/3 ≤ r2 r =2 n n r2 r =1 − n2 ≤ (n3 − 1)/3 ≤ Thus r2 r =1 −1 n 3 (n − 1)/3 ≤ r2 r =1 − 1 ≤ (n3 − 1)/3 + n2 . 3 Dividing by n and allowing n → ∞, we obtain n n −3 r =1 r2 → 1/3. 20 Exercise 1.2.8 If we have the triangular system of size n, one operation is needed to get xn from the nth equation. Substitution of the value of xn in the remaining n − 1 equations reduces them to a triangular system of size n − 1 in about 2(n − 1) operations. We can repeat this process to obtain the complete solution in about n−1 n r =1 2(r − 1) + 1 ≈ 2 r =1 (r − 1) ≈ n2 operations. Exercise 1.2.9⋆ Rather you than me. 21 Exercise 1.2.10 The number of operations is about An3 . I reckon A ≈ 1, so I need about 1000 operations. Reckoning about 100 operations an hour and a 5 hour day gives 2 days. The reader may disagree about everything, but we should still have about the same order of magnitude for the task. 22 Exercise 1.3.5 Row operations Subtract twice first row from second row. 1 −1 3 1 −1 3 → 0 6 −4 252 Divide second row by 6. 1 −1 3 0 6 −4 Add second row to first row. 1 −1 3 0 1 −4 3 → → 1 −1 3 4 0 1 −3 5 10 3 4 0 1 −3 Column operations. Add first column to second. Subtract three times first column from third. 10 0 1 −1 3 → 0 7 −10 252 Divide second column by 7. 10 0 0 7 −10 → 10 0 0 1 − 10 7 Add 10/7 times second column to third. 10 0 0 1 − 10 7 → 100 010 23 Exercise 1.3.9 (i) Subtract 2 times first row from second. Subtract 4 times first row from third. 1 −1 3 1 −1 3 2 5 2 → 0 7 −4 0 7 −4 438 Subtract second row from third. 1 −1 3 1 −1 3 0 7 −4 → 0 7 −4 0 7 −4 00 0 Add first column to second. 1 −1 0 7 00 Subtract 3 times first column from third. 10 0 3 −4 → 0 7 −4 00 0 0 Add 4/7 times second column to third. Divide second column by 7. 100 10 0 0 7 −4 → 0 1 0 000 00 0 (ii) Subtract row. 2 3 4 first row from second and interchange first and second 1 −2 −4 24 5 45 5 2 1 → 1 −2 −4 → 2 4 41 3 41 3 13 Subtract 2 times first row from second and 4 times first row from third. 1 −2 −4 1 −2 −4 2 4 5 → 0 8 13 0 9 19 41 3 Subtract second row from third, interchange second and third. 1 −2 −4 1 −2 −4 1 −2 −4 0 8 13 → 0 8 13 → 0 1 6 0 8 13 01 6 0 9 19 Add twice second row third. 1 0 0 to first and subtract 8 times second row from 10 8 −2 −4 1 6 → 0 1 6 0 0 −35 8 13 24 Divide third row by −35. Subtract 8 times third row from first. 1 10 8 0 1 6 → 0 0 0 0 −35 6 times third row from second and 100 08 1 6 → 0 1 0 001 01 Subtract first row from second and interchange first and second row. 2x + 4y + 5z = −3 3x + 2y + z = 2 4x + y + 3z = 1 2x + 4y + 5z = −3 x − 2y − 4z = 5 4x + y + 3z = 1 x − 2y − 4z = 5 2x + 4y + 5z = −3 4x + y + 3z = 1 Subtract 2 times first row from second and 4 times first row from third. Subtract second row from third, interchange second and third x − 2y − 4z = 5 8y + 13z = −13 9y + 19z = −19 x − 2y − 4z = 5 8y + 13z = −13 y + 6z = −6 x − 2y − 4z = 5 y + 6z = −6 8y + 13z = −13 Add twice second row to first and subtract 8 times second row from third. Divide third row by −35. Subtract 6 times third row from second and 8 times third row from first. x + 8z = −7 y + 6z = −6 −35z = 35 x + 8z = −7 y + 6z = −6 z = −1 x=1 y=0 z = −1 25 Exercise 1.3.10 By subtracting the second row from second, we see that the matrix 111 0 1 1 001 the first and the third from the 1 1 1 has rank 3. By adding the first row to the second and the first row to the third, we obtain a matrix of the same rank 3 with all entries non-zero. 1111 1 2 2 2 . 1122 The same arguments, starting with 1111 1111 0 1 1 1 and 0 0 0 0 , 0000 0000 yields matrices of rank 2 and 1 with all entries non-zero 1111 1111 1 2 2 2 and 1 1 1 1 . 1111 1111 Not possible. All elementary operations (which change a matrix) involve a non-zero row or column which remains non-zero (though possibly moved) after the operation. Thus a rank 0 matrix must be the zero matrix. 26 Exercise 1.4.3 (i) Observe that πi (x + y) + x = (xi + yi ) + zi = xi + (yi + zi ) = πi x + (y + x) . (ii) Observe that πi (x + y) = xi + yi = yi + xi = πi (y + x). (iii) Observe that πi (x + 0) = xi + 0 = xi = πi (x). (v) Observe that πi (λ + µ)x = (λ + µ)xi = λxi + µxi = πi (λx + µy). (vi) Observe that πi (λµ)x = (λµ)xi = λ(µxi ) = πi λ(µy) . (vii) Observe that πi (1x) = 1xi = xi = πi (x) and πi (0x) = 0xi = 0 = πi (0). (viii) Can do as above or x − x = 1x + (−1)x = 1 + (−1) x = 0x = 0. 27 Exercise 2.1.3 If c = 0, then u = (a/c, 0) and v = (0, b/c) are distinct vectors representing points on the line. If c = 0, then suppose, without loss of generality that a = 0. If b = 0, then u = (0, 0) and v = (0, 1) are distinct vectors representing points on the line. If b = 0, then u = (0, 0) and v = (1, −a/b) are distinct vectors representing points on the line. 28 Exercise 2.1.4 (i) Suppose that {v + λw : λ ∈ R} ∩ {v′ + µw : µ ∈ R} = ∅. Then there exist λ0 , µ0 ∈ R such that v + λ0 w = v ′ + µ0 w and so Thus and, similarly so v + λw = v′ + (λ − λ0 + µ0 )w. {v + λw : λ ∈ R} ⊆ {v′ + µw : µ ∈ R} {v′ + µw : µ ∈ R} ⊆ {v + λw : λ ∈ R} {v + λw : λ ∈ R} = {v′ + µw : µ ∈ R}. (ii) The line joining u to u′ is ′ {u + λw : λ ∈ R} with w = u − u . Since w = (σ/τ )v − v′ , the line joining v to v′ is {v + µw : µ ∈ R}. Thus part (ii) follows from part (i). (iii) Observe that 1 1 µ′ u′′ = − ′′ (µu + µ′ u′ ) = (µu + µ′ u′ ) = u + (u′ − u). ′ ′ µ µ+µ µ+µ Exercise 2.1.6⋆ Exercise 2.1.7⋆ 29 Exercise 2.1.9 α α λ = − ⇒ αλ = −α + λα ⇒ λ = . 1−λ β α−β Interchanging α and −(1 − α) and β and −(1 − β ), yields 1−α λ′ = . β−α Exercise 2.2.4⋆ Exercise 2.2.5⋆ 30 Exercise 2.2.8 Let y be the centroid of x1 , x2 , . . . , xq . Observe that q−1 1 1 yj + xj = q q q i=j so y lies on the line joining yj and xj . 1 xi + xj = y, q 31 Exercise 2.2.10 (i) Let y be centre of mass of all the points x1 , x2 , . . . , xq and let M = q=1 mj . j Observe that M − mj 1 mj yj + xj = M M M mi xi + i=j mj xj = y M so y lies on the line joining yj and xj . (ii) We might have m1 + m2 + . . . + mq = 0 and we cannot divide by 0. 32 Exercise 2.3.3 (i) We take the positive square root. (ii) x = 0 ⇒ x (iii) λx 2 = 2 =0⇒ n 2 j =1 (λxj ) n j =1 = λ2 x2 = 0 ⇒ xj = 0 ∀j⇒ x = 0. j n j =1 x2 , so λx = |λ| x . j Exercise 2.3.4⋆ 33 Exercise 2.3.11 The result is trivial if a = b, or b = c, so we assume this is not the case. Setting x = a − b, y = b − c, we see that Theorem 2.3.10 gives a−b + b−c = x + y ≥ x+y = a−c with equality if and only if λ(a − b) = µ(b − c) for some λ, µ > 0 i.e. if and only if a, b, c lie in order along a line. The length of one side of a triangle AB is less than or equal to the sum of the lengths of the other two sides BC and CA with equality if and only if the triangle is degenerate with A, B , C lying in order along a straight line. 34 Exercise 2.3.12 If x lies on the line through a and b then x = ta + (1 − t)b for some t ∈ R. The condition yields that is to say x−a = x−b (1 − t)(a − b) = t(a − b) |1 − t| b − a = |t| b − a so |t| = |1 − t| so t = 1 − t or t = t − 1. The second equation is insoluble so t = 1/2 and x = 1 (a + b). 2 We have proved uniqueness. Direct substitution shows we have a solution. 35 Exercise 2.3.13 (i) Choose X on BC so that AX is perpendicular to BC . By the theorem of Pythagoras and the geometric definition of the cosine and sine |AC |2 = |AX |2 + |CX |2 = |AX |2 + (|BC | − |BX |)2 = |BA|2 sin2 θ + (|BC | − |BA| cos θ)2 = |BA|2 (cos2 θ + sin2 θ) + |BC |2 − 2|BA||BC | cos θ = |BC |2 + |BA|2 − 2|BC | × |BA| cos θ (If the reader is a very careful mathematician, she will not be entirely happy with this argument. She will ask, for example, how we choose the sign of some of the terms. However in this book the algebra is primary and the geometry is illustrative.) (ii) We have a−c 2 = (a − c) · (a − c) = a 2 +c 2 − 2a · c. (iii) The formula in (i) can be rewritten a giving 2 +c 2 − 2 a × c cos θ = a − c a · c = a c cos θ. 2 36 Exercise 2.3.14 Since u·v v·u = , uv uv the angle between v and u is the same as the the angle between u and v If θ is the angle between u and v and φ is the angle between u and −v, then u · (−v) u·v cos φ = =− = − cos θ uv uv and 0 ≤ θ, φ ≤ π , so φ = π − θ 37 Exercise 2.3.16 Let us put the vertices at a = (a1 , a2 , . . . , an ). The diagonals join vertices a and −a (call this diagonal da ). Diagonals da , db are perpendicular if and only if n 0=a·b= ai b i i=1 n Since ai bi = ±1, i=1 ai bi is odd if n is odd and even if n is even. (Proof by induction or modular arithmetic.) Thus if n is odd, a · b = 0 and no diagonals are perpendicular. Now suppose n = 4. In finding the possible angles θ we may suppose without loss of generality that a = (1, 1, 1, 1). The possible angles are given by a·b cos θ = ∈ {r/4 : −4 ≤ r ≤ 4} ab and all the possibilities occur (note that we may get different angles according to the direct assigned to the diagonals). The possible angles are 1 3 0, cos−1 4 , π , cos−1 3 , π , π − cos−1 4 , 6 42 5π , 6 π − cos−1 1 , π 4 the angle 0 and π being the angle between the diagonal and itself and the angle between the diagonal and itself reversed. 38 Exercise 2.3.17 (i) Always true u ⊥ v ⇒ u · v = 0 ⇒ v · u = 0 ⇒ v ⊥ u. (ii) Sometimes false. Take u = w = (1, 0, 0) and v = (0, 1, 0). (iii) Always true u ⊥ u ⇒ u · u = 0 ⇒ u = 0. 39 Exercise 2.3.18 (i) We have u+v 2 = (u + v) · (u + v) =u·u+u·v+v·u+v·v =u 2 +0+0+ v 2 =u 2 +v 2 Consider the right angled triangle OU V with O at 0, U at u and V at v. Then u + v is the length of the hypotenuse. (ii) We have u+v+w 2 = (u + v + w) · (u + v + w) = u · u + v · v + w · w + 2u · v + 2v · w + 2w · u 2 =u +v 2 2 +w +0+0+0= u (iii) If uj ∈ R4 and uj ⊥ uk for k = j , we have 2 4 uj 4 4 = j =1 uj j =1 4 · j =1 k=1 uj · uk 4 j =1 j =1 4 = = uj 4 uj · uj = uj 2 . j =1 2 +v 2 +w 2 40 Exercise 2.4.1 a+b 2 + a−b 2 = (a + b) · (a + b) + (a − b) · (a − b) =( a = 2( a 2 + 2a · b + b 2 ) + ( a 2 + b 2) 2 − 2a · b + b 2 ) The sum of the squares of the lengths of the two diagonals of a parallelogram equals the sum of the lengths of the four sides. 41 Exercise 2.4.2 (i) Consider the parallelogram OACB with vertex 0 at 0, vertex A at a, vertex C at a + b and vertex B at b. The midpoint of the diagonal OC is at 1 x = 1 0 + 2 (a + b) = 1 (a + b). 2 2 The midpoint of the diagonal AB is at 1 1 y = 2 a + 1 b = 2 (a + b). 2 Thus x = y and we are done. The midpoint M of the side CA opposite 0 is at 1 1 m = 2 (a + b) + 1 b = 2 a + b. 2 The point Z given by 1 1 z = 30 + 2m = 3a + 2b 3 3 trisects OM and AB , so we are done. Exercise 2.4.4⋆ 42 Exercise 2.4.6 (i) If (n, m) is a unit vector perpendicular to (u, v ) we have nu = −mv so n = −vx, m = ux and 1 = (u2 + v 2 )x2 = x2 thus (n, m) = (−v, u) or (n, m) = (v, −u). (ii) Let n be a unit vector perpendicular to c Let p = n · a. −1 c as found in (i). Then x = a + tc ⇒ x · n = p and if x · n = p we have (x − a) · n = 0 so x − a = tc for some t. (iii) Let a = pn and let c = u be a unit vector perpendicular to n. x = pn + tu ⇒ x · n = p and, if x · n = p, we have (x − pn) · u = 0 so x − pn = tu for some t. (Or use a geometric argument.) 43 Exercise 2.4.9 We have π1 = {(0, x2 , x3 ) : x2 , x3 ∈ R}, π2 = {(x1 , 0, x3 ) : x2 , x3 ∈ R}, π3 = {(x1 , x2 , x3 ) : x1 + x2 = 1} xj ∈ R}. π1 and π2 meet in the line x1 = x2 = 0. π1 and π3 meet in the line x1 = 0, x2 = 1. π2 and π3 meet in the line x1 = 1, x2 = 0. The lines have no point in common. (i) Now let πj be given by with nj · x = pj , n1 = n2 = n3 = (1, 0, 0) and p1 = −1, p2 = 0, p3 = 1. By inspection no two planes meet. (ii) Finally let πj be given by with nj · x = pj , n2 = n3 = (1, 0, 0), n1 = (0, 1, 0) and p1 = p2 = 0, p3 = 1. We have that π1 and π2 meet in the line given by x1 = x2 = 0 and π1 and π2 meet in the line given by x1 = 1, x2 = 0, but π2 and π3 do not meet. 44 Exercise 2.4.10 {x ∈ R2 : x − a = r} is the singleton {a} if r = 0 and the empty set ∅ if r < 0. 45 Exercise 2.4.11 (i) We have 1 x y(x) = 2 y(x) x y y(x) = 4 1 x = x. x2 2 (ii) Observe that, if a = r, then, if x = 0, setting C = ( a 2 = r2 ⇔ x 2 ⇔y x−a 2 2 − 2x · a + ( a − r2 ) = 0 − r2 ) y ⇔ y − C −1 a = C −1 (C −2 a 2 − r2 ), 2 ↔ 1 − 2y · a + ( a 2 2 − 2y · (C −1 a) + C −1 = 0 − 1) so we transform the circle into a circle centre C −1 a radius (C −1 (C −2 a 2 − 1) 1/2 . (Note that we are taking the square root of a positive number.) (iii) Observe that, if a = r > 0, then, if x = 0, x−a 2 = r2 ⇔ x 2 − 2x · a = 0 ⇔ 1 − 2y · a = 0 so we transform the circle into a line perpendicular to a closest distance to the origin (2 a )−1 . (iv) Goes through without change, replacing ‘circle’ by ‘sphere’ and ‘line’ by plane’ 46 Exercise 2.4.12 x x 2 y − y 2 2 x y y x · − − 2 2 2 x y x y x·x x·y y·y = −2 + x4 x2y2 y4 1 x·y 1 = −2 + 2 2y2 x x y2 x 2 − 2x · y + y 2 = x2y2 = Thus x x 2 2 x−y xy = − y y 2 2 . = x−y . xy Using this result, we have, by the triangle inequality, (if x, y, z = 0) z−x y−z ( x y z )− 1 ( y z − x + x y − z ) = + zx yz y z x z + − − = 2 2 2 z x y z2 x y ≥ − x2 y2 = ( x y z )− 1 z y − x with equality if and only if x −2 x − y are scalar multiples of each other (i.e. x straight line). Thus z x−y ≤ y −2 −2 y and z −2 z − y −2 y , y −2 y z −2 z lie on a z−x + x y if x, y, z = 0. If at least one of the vectors is zero the result is trivial. To obtain the Euclidean result place D at the origin , let A have position vector x, B have position vector y and C position vector z. We now use the notation of Exercise 2.4.11. The condition x −2 x, y −2 y z −2 z are scalar multiples of each other is the same as saying that f (x), f (y), f (y) lie on a straight line so x = f 2 (x), y = f 2 (y) z = f 2 (z) lie on a circle through the origin i.e. A, B , C , D are concyclic. 47 Exercise 3.2.1 zi = aik yk = aik (bkj xj ) = (aik bkj )xj = cij xj 48 Exercise 3.3.4 There are many proofs. We can work coordinatewise e.g. aij + (bij + cij ) = (aij + bij ) + cij so A + (B + C ) = (A + B ) + C Or direct from definition. (A + (B + C ))x = Ax + (B + C )x = Ax + (B x + C x) = (Ax + B x) + C x) = (A + B )x + C x = ((A + B ) + C ))x for all x, so (A + B ) + C = A + (B + C ). 49 Exercise 3.3.6 Suppose C x = x for all x. Taking x to be the column vector with 1 in k th place and zero elsewhere, we get cik = δik . Conversely, if y ∈ Rn n δij yi = yj j =1 so I y = y. 50 Exercise 3.3.9 (iii) By definition. Observe that (B + C )A x = (B + C )(Ax) = B (Ax) + C (Ax) = (BA)x + (CA)x = (BA + CA)x for all x so (B + C )A = BA + CA. (iii) By calculation. We have n n (bij + cij )ajk = j =1 n n bij ajk + cij ajk = j =1 bij ajk + j =1 cij ajk j =1 (iii) By summation convention. Observe that (bij + cij )ajk = bij ajk + cij ajk (iv) By definition. Observe that (λA)B x = (λA)(B x) = λ A(B x) = λ (AB )x = λ(AB ) x for all x so (λA)B = λ(AB ). Again A(λB ) x = A (λB )x = A(λ(B x)) = λ A((B x) = λ (AB )x = λ(BA) x for all x so A(λB ) = λ(AB ). (iv) By calculation. We have n n (λaij )bjk = λ j =1 n aij bjk = j =1 aij (λbjk ) j =1 (iv) By summation convention. Observe that (λaij )bjk = λ(aij bjk ) = aij (λbjk ) (v) By definition. Observe that (IA)x = I (Ax) = Ax and (AI )x = A(I x) = Ax for all x so IA = AI . 51 (v) By calculation. We have n n δij ajk = aik = j =1 n aij δjk = j =1 bij ajk + cij ajk . j =1 (v) By summation convention. Observe that δij ajk = aik = aij δjk . 52 Exercise 3.3.10 We have BA = A ∀A ⇒ BI = I ⇒ B = I. 53 Exercise 3.3.12 If A is an m × n matrix and λ ∈ R, then λA = C where C is the m × n matrix such that λ(Ax) = C x n for all x ∈ R . If A and B are m × n matrices, then A + B = C where C is the m × n matrix such that Ax + B x = C x If A is an m × n matrices and B is an m × p matrix, then AB = C where C is the n × p matrix such that A(B x) = C x for all x ∈ Rn Remark The matrices defined above are certainly unique. The definitions do not by themselves show existence, but the existence may be easily checked. In (i) take cij = λaij . In (ii) take cij = λaij + bij . In (iii) take cij = m air brj . r =1 54 Exercise 3.3.13 (i) Observe that (A + B ) + C )x = (A + B )x + C x = (Ax + B x) + C x = Ax + (B x + C x) = Ax + (B + C )x = ( A + (B + C ) x for all x ∈ Rn so (A + B ) + C = A + (B + C ). (ii) Observe that (A + B )x = Ax + B x = B x + Ax = (B + A)x for all x ∈ Rn so A + B = B + A). (iii) Observe that (A + 0)x = Ax + 0x = Ax + 0 = Ax n for all x ∈ R so A + 0 = A. (iv) Observe that λ(A + B ) x = λ (A + B )x = λ(Ax + B x) = λ(Ax) + λ(B x) = (λA)x + (λB )x = (λA + λB )x n for all x ∈ R so λ(A + B ) = λA + λB . (v) Observe that (λ + µ)A x = (λ + µ)(Ax) = λ(Ax) + µ(Ax) = (λA)x + (µA)x = (λA + µA)x for all x ∈ Rn so (λ + µ) = λA + µA. (vi) Observe that (λµ)A x = (λµ)Ax = λ(µAx) = λ(µA) x for all x ∈ Rn so (λµ)A = λ(µA). (vii) Observe that (0A)x = 0(Ax) = 0 = 0x n for all x ∈ R so 0A = 0. (ix) We have A − A = (1 − 1)A = 0A = 0. 55 Exercise 3.3.14 (i) Observe that (AC )F x = (AC )(F x) = A C (F x) = A (CF )x) = (ACF ) x for all x ∈ Rn , so (AC )F = A(CF ). (ii) Observe that G(A + B ) x = G (A + B )x = G(Ax + B x) = G(Ax) + G(B x) = (GA)x + (GB )x = (GA + GB )x for all x ∈ Rn , so G(A + B ) = GA + GB . (iii) Observe that (A + B )C x = (A + B )(C x) = A(C x) + A(B x) = (AC )x + (BC )x = (AC + BC )x for all x ∈ Rn , so (A + B )C = AC + BC . (iv) Observe that (λA)C )x = (λA)(C x) == λ A(C x) = λ (AC )x) = λ(AC ) x for all x ∈ Rn , so (λA)C = λ(AC ). Observe that A(λC ) x = A (λC )x = A λ(C x) = λ A(C x) = λ (AC )x) = λ(AC ) x for all x ∈ Rn so A(λC ) = λ(AC ). 56 Exercise 3.4.5 The conjecture is false. The matrices A and B given by A= 11 01 and B = are shear matrices, but AB = is not. 21 11 10 11 57 Exercise 3.4.7 Since r = s, we have δrs = 0 and n (δij + λδir δjs )(δjk + µδjr δks ) j =1 n = (δij δjk + λδir δjs δjk + µδjr δks δij + λµδir δjs δjr δks ) j =1 = δik + λδir δks + µδir δks + λµδsr δir δks = δik + (λ + µ)δir δks The result is also obvious geometrically. 58 Exercise 3.5.1 (Lk Lk−1 . . . L1 I )A = Lk Lk−1 . . . L1 A = I, so A is invertible with L k L k − 1 . . . L 1 I = A− 1 . Thus the same set of elementary operations applied in the same order which reduce A to I will transform I to A−1 . 59 Exercise 3.5.2 If we use row operations and column exchanges we can reduce A to I . If we omit column exchanges then the same row operations in the same order produce a matrix with exactly one 1 in each row and column and the remaining entries 0. By exchanging rows we can ensure that the first row is (1, 0, 0, . . . , 0), the second row (0, 1, 0, . . . , 0) and so on, i.e. we can produce I . 60 Exercise 4.1.1 (i) The vertices ABC , BCA, CAB are described in one sense (say anti clockwise) and the same vertices ACB , BAC , CBA in the opposite sense (say clockwise). (ii) (Of course a bit hand waving.) OAXB , 0P QB , BXQ and 0AP anticlockwise but AXP Q clockwise. Writing | area Σ| for the absolute value of the area we have | area OP QB | = | area OAXB | − | area AP QX | + | area OAP | − | area BXQ| = | area OAXB | − | area AP QX |. 61 Exercise 4.1.2 D(a, b) + D(b, a) = D(a + b, b) + D(a + b, a) (By adding second entry to first.) D(a + b, b) + D(a + b, a) = D(a + b, a + b) (Since first entry the same, we can add as shown.) D(a + b, a + b) = D(a + b − (a + b), a + b) (By subtracting second entry from first entry.) D(a + b − (a + b), a + b) = D(0, a + b) (Just do the calculation.) D(0, a + b) = 0 (Area degenerate parallelogram.) D(a, b) is the area of the parallelogram with vertices 0, a, a + b, b described in that order. D(b, a) is the area of the same parallelogram but with the vertices laid out in the opposite sense 0, b, a + b, a. 62 Exercise 4.1.3 a1 b ,1 a2 b2 D a1 b ,1 0 b2 =D 0 b ,1 a2 b2 +D (Since second column the same, can add first columns.) a1 b ,1 0 b2 D 0 b ,1 a2 b2 +D b1 b a1 , 1− b2 0 a1 =D a1 0 0 b ,1 a2 b2 +D − b2 a2 0 a2 (Since we may subtract multiples of the first column from the second.) a1 b ,1 0 b2 D − b1 a1 a1 0 +D a1 0 , b2 0 =D 0 b ,1 a2 b2 − b2 a2 0 a2 0 b ,2 a2 0 +D (Just doing the calculation.) D a1 0 , b2 0 =D 0 b ,2 a2 0 +D a1 0 , b2 0 −D b2 0 , a2 0 (Interchanging columns.) D a1 0 , b2 0 −D 0 b2 , a2 0 = (a1 b2 − a2 b1 )D 0 1 , 1 0 = a1 b 2 − a2 b 1 . (Using the rules D(λa, b) = D(a, λb) = λD(a, b) and the fact that that the area of a unit square is 1.) 63 Exercise 4.2.1 (Not a good idea, but possible as follows.) D(AB ) = D a11 b11 + a12 b21 a11 b12 + a12 b22 a21 b11 + a22 b21 a21 b12 + a22 b22 = (a11 b11 + a12 b21 )(a21 b12 + a22 b22 ) − (a11 b12 + a12 b22 )(a21 b11 + a22 b21 ) = a11 a22 (b11 b22 − b12 b21 ) − a12 a21 (b11 b22 − b12 b21 ) = D(A)D(B ). 64 Exercise 4.2.2 (i) We have DI = D 10 01 = 1 × 1 − 0 × 0 = 1. The area of a unit square is 1. (i) We have 01 10 DE = D and E x y = 01 10 = 0 × 0 − 1 × 1 = −1, x y 0x + 1y 1x + 0y = = y . x We observe that E cos t sin t = sin t cos t = cos( π − t) 2 sin( π − t) 2 runs clockwise from (0, 1) back to (0, 1) as (cos t, sin t)T runs anticlockwise from (1, 0) back to (1, 0). (iii) We have D 1λ 01 =1×1−λ×0=1 D 10 λ1 = 1 × 1 − 0 × λ = 1. D a0 0b = a × b − 0 × 0 = ab and (iv) We have corresponding to the fact that a rectangle with sides of length a and b has area ab. Exercise 4.2.3⋆ 65 Exercise 4.3.2 As we hoped, ǫ11 a11 a12 + ǫ12 a11 a22 ǫ21 a21 a12 + ǫ22 a21 a22 = a11 a22 − a21 a12 . 66 Exercise 4.3.3 The result is automatic if r, s and t are not distinct. When r, s andt are distinct, we check the case r = 1 when either s = 2, t = 3 or s = 3, t = 2 in Definition 4.3.1. We now do the same for r = 2 and r = 3. 67 Exercise 4.3.10 Observe that writing C = AB D = C T , A′ = AT , B ′ = B T m dij = cji = m m a′rj b′ir ajr bri = r =1 r =1 T b′ir a′rj = r =1 T T (for 1 ≤ i ≤ p, 1 ≤ j ≤ n). Thus (AB ) = B A . 68 Exercise 4.3.13 x → Er,s,λ x is a shear which leaves volume unchanged. Let D have ith diagonal entry di . x → Dx (where D is a diagonal matrix) stretches by |di | in the 0xi direction and reflects if dii < 0. Thus the volume is multiplied by n=1 di = det D. i x → P (σ )x exchanges the handedness of coordinates (or equivalently is a reflection in a particular plane) which multiplies volume by −1. If A is any 3 matrix then A = a1 A2 . . . Ak with the Aj elementary matrices. Since Ax = A1 (A2 (A3 . . . (Ak x) . . .)) the transformation x → Ax rescales by k j =1 det Aj = det A. 69 Exercise 4.3.14 The map x → M x changes the length scale by λ and the volume scale by λ3 . (This is just a special case of the map x → Dx considered in the previous question.) 70 Exercise 4.3.15 Let F (r, s, t) = ǫijk air ajs akt . We observe that F (r, s, t) is the determinant of the matrix with first row the rth row of A, second row the sth row of A and third row the tth row of A. so Thus interchanging any two of r, s, t multiplies F (r, s, t) by −1 and F (r, s, t) = ǫrst K for some constant K . Since F (1, 2, 3) = det A we have ǫijk air ajs akt = F (r, s, t) = ǫrst det A. (ii) Writing C = AB , we have ǫrst det AB = ǫrst det C = ǫijk cir cjs ckt = ǫijk aiu bur ajv bvs akw bwt = ǫijk aiu ajv akw bur bvs bwt = ǫuvw det Abur bvs bwt = ǫrst det det B Taking r = 1, s = 2, t = 3 we get det AB = det A det B . Exercise 4.3.16⋆ 71 Exercise 4.4.1 There is no non-trivial χ. Observe that so χijkrs = 0. χijkrs = −χkijrs = χjkirs = −χijkrs , 72 Exercise 4.4.4 We have (σ 2 − σ 1)(σ 3 − σ 1)(σ 4 − σ 1)(σ 3 − σ 2)(σ 4 − σ 2)(σ 4 − σ 3) ζ (σ ) = , (2 − 1)(3 − 1)(4 − 1)(3 − 2)(4 − 2)(4 − 3) ζ (τ ) = ζ (ρ) = (3 − 2)(1 − 2)(4 − 2)(1 − 3)(4 − 3)(4 − 1) = 1, (2 − 1)(3 − 1)(4 − 1)(3 − 2)(4 − 2)(4 − 3) (3 − 2)(4 − 2)(1 − 2)(4 − 3)(1 − 3)(1 − 4) = −1. (2 − 1)(3 − 1)(4 − 1)(3 − 2)(4 − 2)(4 − 3) 73 Exercise 4.4.6 (i) If t ∈ {1, 2, i} / αρα1 = αρ2 = α1 = i = τ 1 αρα2 = αρ1 = α2 = 2 = τ 2 αραi = αρi = αi = 1 = τ i αραt = αρt = αt = t = τ t (ii) If j = 1, then the result follows from part (iv) of the lemma. If not, let τ be as in part (iv) of the lemma and β ∈ Sn interchange 1 and j leaving the remaining integers unchanged. Then, by inspection, βτ β (r) = κ(r) for all 1 ≤ r ≤ n and so βτ β = κ. It follows that ζ (κ) = ζ (β )ζ (τ )ζ (β ) = ζ (β )2 ζ (τ ) = −1. 74 Exercise 4.4.7 If all the suffices i, j , k , l are distinct, then σ (1) = i, σ (2) = j , σ (3) = k , σ (4) = l gives a unique σ ∈ S4 , so ǫijkl = ǫσ(1)σ(2)σ(3)σ(4) = ζ (σ ) does define ǫijkl . If τ interchanges two suffices and leaves the rest unchanged ǫτ (i)τ (j )τ (k)τ (l) = ǫτ σ(1)τ σ(2)τ σ(3)τ σ(4) = ζ (τ σ ) = ζ (τ )ζ (σ ) = −ζ (σ ) = −ǫijkl If i, j , k , l are not all distinct and i′ , j ′ , k ′ , l′ is some rearrangement ǫijkl = 0 = −ǫi′ j ′ k′ l′ . Finally ǫ1234 = ζ (ι) = 1. 75 Exercise 4.4.8 Observe that 1−1 = 1 and (−1)−1 = −1. Since ζ (σ )ζ (σ −1 ) = ζ (σσ −1 ) = ζ (ι) = 1, it follows that ζ (σ ) = ζ (σ −1 ). Thus det AT = ζ (σ )a1σ(1) a2σ(2) . . . anσ(n) σ ∈Sn = ζ (σ )aσ−1 (1)1 aσ−1 (2)2 . . . aσ−1 (n)n σ ∈Sn ζ (σ −1 )aσ−1 (1)1 aσ−1 (2)2 . . . aσ−1 (n)n = σ ∈Sn = ζ (τ )aτ (1)1 aτ (2)2 . . . aτ (n)n τ ∈Sn = det A 76 Exercise 4.4.9 (i) We have det 11 xy = y − x. (ii) Each term ζ (σ )aσ(1)1 aσ(2)2 aσ(3)3 in the standard expansion is a multinomial of degree 3. 111 F (x, x, z ) = det x x z = 0, x2 x2 z 2 because two columns of the matrix are identical. Thus y − x must be be a factor of F (x, y, z ). Similarly z − y and z − x are factors. Since F has degree 3 and (y − x)(z − y )(z − x) has degree 3, F (x, y, z ) = A(y − x)(z − y )(z − x) for some constant A. By considering the terms in the standard determinant expansion, we know that the coefficient of yz 2 is 1. Thus A = 1 and F (x, y, z ) = (y − x)(z − y )(z − x). (iii) Since each term in the standard expansion of det V is a multinomial of degree 0 + 1 + . . . + (n − 1) = (n − 1)n/2, F is a multinomial of degree (n − 1)n/2. F (x1 , x1 , x3 , x4 , . . . , xn ) = 0 because two columns of the associated matrix are identical. Thus x2 − x1 must be be a factor of F . Similarly xj − xk is a factor for each j > k . Since F has degree (n − 1)n/2 and i>j (xi − xj ) has the same degree, F (x1 , x2 , . . . , xn ) = A i>j (xi − xj ) for some constant A. By considering the terms in the standard determinant expansion, we j know that the coefficient of n=2 xj −1 is 1. Thus A = 1 and j F (x1 , x2 , . . . , xn ) = i>j (xi − xj ). (iv) We have F (xσ(1) , xσ(2) , . . . , xσ(n) ) ˜ = ζx (σ ) = F (x1 , x2 , . . . , xn ) i>j xσ(i) − xσj . xi − xj 77 Now i>j (xσ(i) − xσj ) = Bσ i>j (xi − xj ) where B depends only on σ (and not on x). Thus ˜ ˜ ζx (σ ) = ζy (σ ) with yj = j and so ˜ Thus ζ = ζ . ˜ ζx (σ ) = ζ (σ ). 78 Exercise 4.5.1 Each of the n! terms involves n multiplications (in addition to finding the appropriate sign) so we need n × n! multiplications. We can either use a Stirling approximation or an electronic calculator (which probably uses some version of Stirling’s formula) to obtain an estimate of about 36 000 000 for n = 10 and about 4.9 × 1019 for n = 20. 79 Exercise 4.5.3 If A = (aij ) is an n × n matrix which is both upper and lower triangular then aij = 0 if i < j or if j < i so A is diagonal.Each of the n! terms involves n multiplications (in addition to finding the appropriate sign) so we need nn! multiplications. We can either use a Stirling approximation or an electronic calculator (which probably uses some version of Stirling’s formula) to obtain an estimate of about 36 000 000 for n = 10 and about 4.9 × 1019 for n = 20. 80 Exercise 4.5.3 If A = (aij ) is an n × n matrix which is both upper and lower triangular then aij = 0 if i < j or if j < i so A is diagonal. 81 Exercise 4.5.5 (i) By row and column operations on the first r rows and columns we can reduce A′ 0 A0 to C ′ = C= 0B 0B ′ ′ with det C = K det C , det A = K det A and A′ lower triangular. By row and column operations on the last s rows and columns we can reduce C ′ to C ′′ with A′ 0 C ′′ = 0 B′ with det C ′′ = K ′ det C ′ , det B ′ = K ′ det B and B ′ lower triangular. We now have det C = KK ′ det C ′′ = KK ′ det A′ det B ′ = det A det B. (ii) FALSE. Consider 0 0 F = 1 0 With the suggested notation 1 0 0 0 0 1 0 0 0 0 0 1 det A = det B = det C = det D = 0 but det F = 1 = 0 = det A det D − det B det C. 82 Exercise 4.5.6 Dividing the first row by 24 det 3 1 52 2, 123 6 2 = 2 det 3 1 2 . 523 3 Subtracting multiples of the first row from 1 123 3 1 2 = 2 det 0 2 det 0 523 the second and third row, 2 3 −5 −7 . −8 −12 Expanding by first row, 12 3 −5 −7 . 2 det 0 −5 −7 = 2 det −8 −12 0 −8 −12 Dividing the first row by −1 and the second row by −4, −5 −7 −8 −12 = 8 det 52 21 = 8 det 57 . 23 Subtracting two times the second row from the first (hardly necessary, we could have finished the calculation here), 8 det 10 . 21 From the definition, 8 det 10 21 = 8. 83 Exercise 4.5.7 (i) Observe that T a11 a12 a13 a11 a12 a13 det a21 a22 a23 = det a21 a22 a23 a31 a32 a33 a31 a32 a33 3 = aiσ(i) σ ∈S3 i=1 = a11 a22 a33 − a11 a23 a32 + a12 a23 a31 − a12 a21 a33 + a13 a21 a32 − a13 a22 a31 = a11 det a22 a23 a32 a33 − a12 det a21 a23 a31 a33 + a31 det a21 a22 . a31 a32 (ii) We have 246 31 32 12 + 6 det − 4 det det 3 1 2 = 2 det 52 53 23 523 = 2 × (−1) − 4(−1) + 6 × 1 = 8. 84 Exercise 4.5.8 (i) We look at each of the four terms ar1 det Br1 say. If r = 1, j then det Br1 changes sign to give −ar1 det Br1 . a11 det B11 changes to −aj 1 det Bj 1 and aj 1 det Bj 1 changes to −a11 det B11 . Thus ˜ F (A) = −F (A). (iii) The result is already proved if i or j takes the value 1. If not, interchanging row 1 with row i, then interchanging row 1 of the new matrix with row j of the new matrix and finally interchanging row 1 ˜ and i again transforms A to A. By part (i) ˜ F (A) = −F (A). If rows i and j are the same this gives F (A) = −F (A) and so F (A) = 0. (iv) By inspection ˜ F (A) = F (A) + F (B ) where B is the matrix A with the first row replaced by the ith row. By part (iii), F (B ) = 0 so ˜ F (A) = F (A). By considering the effect of interchanging the first and j th row, we get the more general result ¯ F (A) = F (A). (v) If we now carry out the diagonalisation procedure of Theorem 3.4.8 on A using the rules above and observe that we get F (I ) = 1 = det I, F (A) = det A. 85 Exercise 4.5.10 (i) The argument goes through essentially word for word (replacing 4 by n). The formula n a1j A1j = det A. j =1 is the row expansion formula so obtained. (ii) If i = 1 there is nothing to prove. If i = 1, we argue as follows. If B is the matrix obtained from A by interchanging row 1 and row i. n det A = − det B = − (iii) If i = k , then n b1j B1j = j =1 aij Aij . j =1 n aij Akj = det C j =1 where C is an n × n matrix with ith and k th rows the same. Thus n aij Akj = 0. j =1 (iv) Conditions (ii) and (iii) together give n aij Akj = δkj det A j =1 86 Exercise 4.5.13 We have so det A−1 det A det A−1 = det AA−1 = det I = 1 = (det A)−1 . 87 Exercise 4.5.14 n The formula of Exercise 4.5.10 (v) shows that k=1 bk Akj is the determinant of a matrix obtained from A by replacing the j th column of A by b. Thus n n aij det Bj = j =1 n aij j =1 n bk Akj k=1 n = aij bk Akj j =1 k=1 n n aij bk Akj = k=1 j =1 n n = aij Akj bk k=1 n = j =1 bk δik det A = bi det A k=1 Thus, if det A = 0, det Bj det A gives a solution of Ax = b (and, since there is only one solution, it is the solution). xj = 88 Exercise 4.5.15 The statement is true but you still need to find the determinant of two n × n matrices. Unless we deal with very small systems of equations (corresponding to a 3 × 3 matrix, say) the labour involved in computing det A (at least by the methods given in this book or any I am aware of) is, at best, comparable with the effort of finding all the solutions of Ax = b 89 Exercise 4.5.16 (i) Observe that, since Sn is a group n perm AT = aiσ(i) σ ∈Sn i=1 n = aσ−1 (i)i σ ∈Sn i=1 n aτ (i)i = perm A = σ ∈Sn i=1 (ii) Both statements false. If we set A= 11 1 −1 11 , B= 11 then perm A = 4 = 0, but det A = 0, and det B = −2, but perm B = 0. (iii) Write Aji for the n − 1 × n − 1 matrix obtained by removing the ith row and j th column from A. We have n perm A = a1i perm Ai1 i=1 (iv) By (iii), perm A ≤ nK max Ai1 i so, by induction, | perm A| ≤ n!K n . If we take A(n) to be the n × n matrix with all entries K (where K ≥ 0), then perm A(n) = nK perm A(n − 1) so, by induction, perm A(n) = n!K n . (v) If B is the n × n matrix with bij = |aij | | det A| ≤ perm B ≤ n!K n . Hadamard’s inequality is proved later. 90 Exercise 4.5.17 (i) True. If A is antisymmetric, and det A = b2 = 0 if A = 0. 0b , −b 0 (ii) False. Consider (iii) True. 0 −1 0 0 1 0 0 0 0 0 0 0 0 0 . 0 0 det A = det AT = det(−A) = (−1)n det A = − det A. 91 Exercise 5.1.1 If we work over Z we cannot always divide. Thus the equation 2x = 1 has no solution, although the 1 × 1 matrix (2) has non-zero determinant. Exercise 5.1.2⋆ 92 Exercise 5.2.7 I would be inclined to pick (iv) and (vi). (ii) (f + g )(x) = f (x) = g (x) = g (x) + f (x) = (g + f )(x) for all x so f + g = g + f. (iii) (f +0)(x) = f (x)+0(x) = f (x)+0 = f (x) for all x so f +0 = f . (iv) (λ(f +g ))(x) = λ((f +g )(x)) = λ(f (x)+g (x)) = λf (x)+λg (x) = (λf )(x) + (λg )(x) = (λf + λg )(x) for all x so λ(f + g ) = λf + λg . (v) ((λ + µ)f )(x) = (λ + µ)(f (x)) = λf (x) + µf (x) = (λf )(x) + (µg )(x) + (λf + µg )(x) for all x so (λ + µ)f = λf + µf . (vi) ((λµ)f )(x) = (λµ)(f (x)) = (λµ)f (x) = λ(µf (x)) = λ((µf )(x)) = (λ(µf ))(x) for all x so (λµ)f = λ(µf ). (vii) (1f )(x) = 1 × f (x) = f (x) and (0f )(x) = 0 × f (x) = 0 = 0(x) for all x and so 1f = f and 0f = 0. 93 Exercise 5.2.8 The correspondence FX ↔ Fn given by f ↔ (f (1), f (2), f (3), . . . , f (n)) identifies FX with the known vector space Fn . 94 Exercise 5.2.11 Observe that all these sets are subsets of the vector space RR so we may use Lemma 5.2.10. (i) Subspace. If f and g are 3 times differentiable so is λf + µg . (ii) Not a vector space. Let f (t) = t2 . Then f is in the set but (−1)f is not. (iii) Not a vector space. Let P (t) = t. Then P is in the set but (−1)P is not. (iv) Subspace. If P , Q in set, then λP + µQ is a polynomial and (λP + µQ)′ (1) = λP ′ (1) + µQ′ (1) = 0 (v) Subspace. If P , Q in set, then λP + µQ is a polynomial and 1 1 (λP (t) + µQ(t) dt (λP + µQ)(t) dt = 0 0 1 =λ 1 P (t) dt + µ 0 q (t) dt = 0. 0 (vi) Not a vector space. Let h(t) = max{(1 − 10|t|), 0} and observe that h(t)3 dt = A = 0. If f (t) = h(t) − h(t + 1/5) and g (t) = h(t) − h(t + 2/5), then 1 1 f (t) dt = −1 −1 but g (t) dt = A − A = 0 1 −1 (f (t) + g (t)) dt = 8A − 2A = 0. (vii) Not a vector space. If f (t) = −g (t) = t3 then f and g have degree exactly 3 but f + g = 0 does not. (viii) Subspace. If P , Q in set, then so is λP + µQ. 95 Exercise 5.3.2 T (0) = T (00) = 0T (0) = 0. 96 Exercise 5.3.3 Let f, g ∈ D (i) We have δ (λf + µg ) = (λf + µg )(0) = λf (0) + µg (0) = λδf + µδg. (ii) We have D(λf + µg ) = (λf + µg )′ = λf ′ + µg ′ = λDf + µDg. (iii) We have K (λf + µg ) (x) = (x2 + 1)(λf + µg )(x) = (x2 + 1)(λf (x) + µg (x)) = λ(x2 + 1)f (x) + µ(x2 + 1)g (x) = λ(Kf )(x) + µ(Kg )(x) = (λKf + µKg )(x) for all x and so K (λf + µg ) = λKf + µKg. (iv) We have x (λf (t) + µg (t)) dt J (λf + µg ) (x) = 0 x =λ x f (t) dt + µ 0 g (t) dt = (λJf + µJg )(x) 0 for all x and so J (λf + µg ) = λJf + µJg. 97 Exercise 5.3.10 (i) We have ι(λx + µy) = λx + µy = λιx + µιy so ι is linear (ii) Just a restatement of definitions. (iii) The fundamental theorem of the calculus states that DJ = ι. However JD1 = J 0 = 0 = 1. To see that J is injective observe that Jf = Jg ⇒ DJf = DJg ⇒ f = g However Jf (0) = 0 so 1 ∈ J D so J is not surjective. / To see that D is surjective observe that D(Jf ) = f . However D0 = D1 = 0 so D is not injective. (iv) Observe that (αβ )(β −1 α−1 ) = α(ββ −1 )α−1 = αια−1 = α−1 α = ι and similarly (β −1 α−1 )(αβ ) = ι 98 Exercise 5.3.11 The only if part is trivial To prove the if part suppose Then T u = 0 ⇒ u = 0. Tx = Ty ⇒ Tx − Ty = 0 ⇒ T (x − y) = 0 ⇒x−y =0 ⇒ x = y. 99 Exercise 5.3.14 (i) If α, β ∈ GL(U ), then α and β are bijective, so αβ is, so αβ ∈ GL(U ). (ii) Observe that if α, β, γ ∈ GL(U ), then (αβ )γ u = (αβ )(γ u) = α β (γ u) = α β (γ u) = α (βγ )u = α(βγ ) u for all u ∈ U so (αβ )γ = α(βγ ). (iii) ι ∈ GL(U ) and αι = ια = α for all α ∈ GL(U ). (iv) Use the definition of GL(U ). Let α(x, y ) = (x, 0), β (x, y ) = (y, x). βα(x, y ) = (0, x). (Could use matrices.) Then αβ (x, y ) = (y, 0), 100 Exercise 5.3.16 (i) Not a subgroup. Observe that if A= 11 20 then A is invertible (since det A = 0) but A2 = 31 . 22 (ii) Yes, a subgroup since det ι = 1 > 0, and det α, det β > 0 ⇒ det αβ = det α det β > 0 det α > 0 ⇒ det α−1 = (det α)−1 > 0. (iii) Not a subgroup. If α = 2ι then det α = 2n ∈ Z but (if n ≥ 2) det α−1 = 2−n ∈ Z. / (iv) Yes, a subgroup. If α, β ∈ H4 with matrices A and B then AB has integer entries and det AB = det A det B = 1. Further since A−1 = (det A)−1 Adj A = Adj A and Adj A has integer entries, α−1 ∈ H4 . Finally ι ∈ H4 . (v) Yes, a subgroup. Let Sn be the group of permutations σ : {1, 2, . . . , n} → {1, 2, . . . , n}. Let ασ be the linear map whose matrix (δi,σi ) (that is to say (aij ) with − aiσi = 1, aij = 0 otherwise). Then αι = ι, ασ ατ = ατ σ and ασ 1 = ασ−1 . 101 Exercise 5.4.3 Observe that n j =1 n λj (ej + y) = 0 ⇒ ⇒ n λj ej + j =1 n λ j ak e k = 0 j =1 k=1 n n λj ej + j =1 λ k aj e j = 0 j =1 k=1 n ⇒ n n λ j + aj λr ej = 0 r =1 j =1 n ⇒ λ j + aj Thus, if r =1 λr = 0 ∀j n λj ej + y = 0 j =1 and we write K = − n r =1 λr , λj = Kaj . Summing we obtain n −K = n λj = K j =1 aj j =1 so, a1 + a2 + . . . + an + 1 = 0, K = 0 and λj = 0 showing that the vectors e1 + y are linearly independent. If a1 + a2 + . . . + an + 1 = 0, then n aj (ej + y) = 0 j =1 and (since y = 0) not all the aj are zero, so we do not have linear independence. 102 Exercise 5.4.11 U and V are the null spaces of linear maps from R4 to R2 . The elements of U ∩ V are given by x + y − 2z + t = 0 −x + y + z − 3t = 0 x − 2y + z + 2t = 0 y + z − 3t = 0. Subtracting the 4th equation from the 2nd yields x = 0 so the system becomes y − 2z + t = 0 x=0 −2y + z + 2t = 0 y + z − 3t = 0. Adding the 1st and 3rd equation reveals that the 4th equation is superfluous so the system is y − 2z + t = 0 x=0 −2y + z + 2t = 0, or equivalently x=0 y − 2z + t = 0 −3z + 6t = 0 so x = 0, z = 2t, y = −2t. Thus U ∩ V has basis e1 = (0, −2, 2, 1)T . The equations for U yield x + y − 2z + t = 0 2y − z − 2t = 0 so by inspection e2 = (1, 1, 2, 0)T is in U . Since e1 and e2 are linearly independent and dim U = 2 we have basis for U (To see that dim U = 2 either quote general theorems or observe that x and y determine z and t.) The equations for v yield x − 2y + z + 2t = 0 y + z − 2t = 0 103 so, by inspection, e3 = (3, 1, 2, 0)T is in U . Since e1 and e3 are linearly independent and dim V = 2 (argue as before) we have basis for V . The proof of Lemma 5.4.10 now shows that e1 , e2 form a basis for e3 104 Exercise 5.4.12 (i) Since U ⊇ V + W ⊇ V, W we have dim U ≥ dim(V + W ) ≥ dim V, dim W By Lemma 5.4.10 dim(V + W ) = dim V + dim W − dim(V ∩ W ) ≥ dim V + dim W. Putting these two results together, we get min{dim U, dim V + dim W } ≥ dim(V + W ) ≥ max{dim V, dim W }. (ii) Consider a basis e1 , e2 , . . . , en for U . Let V = span{e1 , e2 , . . . , er } Then and W = span{et−s+1 , et−s+2 , . . . , et } V + W = span{e1 , e2 , . . . , et } dim V = r, dim W = s, dim(V + W ) = t. 105 Exercise 5.4.13 We use row vectors. E is the null space of a linear map from R3 to R so a subspace. Let e1 = (1, −1, 0), e2 = (1, 0, −1). We have e1 , e2 ∈ U . x1 e1 + x2 e2 = 0 ⇒ (x1 + x2 , −x1 , −x2 ) = 0 ⇒ x1 = x2 = 0 so e1 , e2 are independent. If x ∈ U then x = (x1 , x2 , x3 ) ⇒ x = (−x2 − x3 , x2 , x3 ) ⇒ x = x2 e1 + x3 e2 . Thus e1 , e2 span U and so form a basis. I do not think everybody will choose the basis and so there cannot be a ‘standard basis’ in this case. 106 Exercise 5.4.14 (i) 0 ∈ V for all V ∈ V and so 0 ∈ V ∈V V. Further, if λ, µ ∈ F, u, v ∈ V ∈V V ⇒ u, v ∈ V ∀V ∈ V ⇒ λu + µv ∈ V ∀V ∈ V ⇒ λu + µv ∈ Thus V ∈V V. V ∈V V is a subspace of U . (ii) By (i) W is a subspace of U . Since E ⊆ V for all V ∈ V , we have E ⊆ W . If W ′ is as stated W ′ ∈ V so W ′ ⊇ W . (iii) Let n W′ = j =1 λj ej : λj ∈ F for 1 ≤ j ≤ n . Observe that E ⊆ W ′ (take λj = δij ). and W is a subspace of U since n n n λ λj ej + µ j =1 µj ej = j =1 (λλj + µµj ej . j =1 Any subspace of U containing E must certainly contain W ′ . Thus W = W ′. 107 Exercise 5.5.3 Always false. 0 ∈ α−1 v. / 108 Exercise 5.5.5 We use the rank-nullity theorem. α injective ⇔ α−1 (0) = 0 ⇔ dim α−1 (0) = 0 ⇔ n − dim α−1 (0) = n ⇔ dim α(U ) = n ⇔ α surjective We now note that bijective means injective and surjective and that a map is bijective if and only if it is invertible. 109 Exercise 5.5.6 Adding the second and third row to the top row, we get a + 2b a + 2b a + 2b abb a b det b a b = det b b b a bba 1 1 1 111 0 = (2a + b)(a − b)2 . = (a + 2b) b a b = (2a + b) 0 a − b 0 0 a−b bba Thus, if a = b and a = −2b, α has rank 3 α−1 (0) = 0, has empty basis and α(U ) = U has basis (1, 0, 0)T , (0, 1, 0)T , (0, 0, 1)T . If a = b, there are two cases. If a = 0, α has rank 3 α(U ) = 0 has empty basis and α−1 (0) = U has basis (1, 0, 0)T , (0, 1, 0)T , (0, 0, 1)T . If a = b = 0, then (x, y, z ) ∈ α−1 (0) ⇔ x + y + z = 0 so α has rank 3 − 2 = 1 α−1 (0) has basis (1, 0, −1)T , (0, 1, −1)T . U has basis (1, 0, −1)T , (0, 1, −1)T , (0, 0, 1)T , so α(U ) has basis α(0, 0, 1)T = (a, a, a)T so basis (1, 1, 1)T . If a = −2b = 0, then (x, y, z ) ∈ α−1 (0 ⇔ x − 2y + z = 0, x + y − 2z = 0 ⇔ x = y = z so α has rank 3 − 1 = 2 α−1 (0) has basis (1, 1, 1)T . U has basis (1, 1, 1)T , (0, 1, 0)T , (0, 0, 1)T so α(U ) has basis α(0, 1, 0)T = (b, a, b)T α(0, 1, 0)T = (b, b, a)T so basis (1, −2, 1)T , (1, 1, −2). Exercise 5.5.12⋆ 110 Exercise 5.5.13 (i) If A is an m × n matrix with rows a1 , a2 , . . . am then the row rank of A is dimension of span{a1 , a2 , . . . , am }. (ii) If B is a non-singular m × m matrix then if x is a row vector with m entries so xB = 0 ⇒ xBB −1 = 0B −1 ⇒ x = 0 ⇒ xB = 0 x = 0 ⇔ xB = 0. It follows that m m j =1 λj aj B = 0 ⇔ ⇔ λj aj B=0 j =1 m λj aj = 0, j =1 so dim span{a1 B, a2 B, . . . , am B } = dim span{a1 , a2 , . . . , am } and the row rank of A and AB are the same. Since the elementary column operations correspond to post multiplication by invertible matrices, the row rank is unaltered by elementary column operations. A similar argument with pre-multiplication replacing post-multiplication shows that row rank is unaltered by elementary row operations. Similarly the column rank is unaltered by elementary row and column operations. (iii) By elementary row and column operations any matrix A can be transformed to a matrix C = (cij ) with cij = 0 if i = j . By inspection, C has row and column rank equal to the number of non-zero entries cii . Thus, by (ii), the row rank and column rank of A are the same. 111 Exercise 5.5.14 If a = b = 0, then r = 0. If a = 0 and b = 0, the matrix is a0 0 a a 0 00 0 0 0 0 0 0 0 a The first and third rows are identical, so r ≤ 3. However, the matrix obtained by removing the third row and first column a00 0 a 0 00a is non-singular, so r = 3. If a = 0, b = 0, we get 0 0 0 0 0 0 0 b b 0 0 0 0 b . b 0 The second and third rows are identical, so r ≤ 3. However, the matrix obtained by removing the third row and first column 0b0 0 0 b b00 is non-singular (for example, because the determinant is non-zero), so r = 3. If a = b = 0, the matrix is a 0 a 0 0 a 0 a a 0 0 0 0 a . a a The second and fourth rows are identical so r ≤ 3. However, the matrix obtained by removing the fourth row row and column a0a 0 a 0 00a is non-singular (for example, because the determinant is non-zero) so r = 3. 112 If a, b = 0 and a = b, then subtracting the first row from the third row and the a/b times the second row from the fourth a0 b 0 a0 b 0 a0b0 0 a 0 b b → 0 a 0 b → 0 a 0 0 0 −b 0 0 −b b a 0 0 b b 2 0 0 0 a − b /a 0b 0 a 0b0a yields a non-singular matrix (e.g. by looking at the determinant), so r = 4 and the original matrix was non-singular. 113 Exercise 5.5.16 Consider the linear maps α and β whose matrices with respect to a given basis are (aij ) with aii = 1 if 1 ≤ i ≤ r, aij = 0, otherwise, bii = 1 if r − t + 1 ≤ i ≤ r + s − t, bij = 0, otherwise. (Note that we use the facts that r, s ≥ t and r + s − t ≤ n.) Then AB has matrix (cij ) with cii = 1 if r − t + 1 ≤ i ≤ r, cij = 0, otherwise. Then α has rank r, β rank s and γ rank t. 114 Exercise 5.5.17 By noting that (1, −1, 0, 0, . . . , 0)T , (1, 0, −1, 0, . . . , 0) . . . , (1, 0, 0, 0, . . . , −1)T are n − 1 linearly independent eigenvectors with eigenvalue 0 and (1, 1, . . . , 1)T is an eigenvector with eigenvalue n for the n × n matrix A with all entries 1, we see that the characteristic polynomial of A is det(tI − A) = tn−1 (t − n). Thus A + sI is invertible if and only if s = 0, −n. Let P be as stated. Then P P T = Q with n qij = pir pjr = r =1 1 k if i = j if i = j T Thus P P = A + (k − 1)I and, since k ≥ 2 we have P P T invertible so P is invertible so of full rank n. If k = 1, then there can only be one party P = (1 1 1 . . . 1)T and has rank 1. The proof fails because P P T = A is not of full rank. 115 Exercise 5.5.18 (i) Observe that (α + β )U ⊆ V , so dim V ≥ dim(α + β )U . Since (α + β )u = αu + β u ∈ αU + βU , we have (α + β )U ⊆ αU + βU , so, by Lemma 5.4.10, dim(α + β )U ≤ dim(αU + βU ) ≤ dim αU + dim βU. Thus min{dim V, rank α + rank β } ≥ rank(α + β ). (ii) By (i), rank(α+β )+rank β = rank(α+β )+rank(−β ) ≥ rank (α+β )−β = rank α. Thus and Thus rank(α + β ) ≥ rank α − rank β rank(α + β ) = rank(β + α) ≥ rank β − rank α. rank(α + β ) ≥ | rank α − rank β |. (iii) Since α + β = β + α, there is no loss of generality in assuming r ≥ s. Fix bases for U and V . If t ≥ r, consider the linear map α given by the matrix (aij ) with aii = 1 if 1 ≤ i ≤ r, aij = 0 otherwise, and the linear map β given by the matrix (bij ) with bii = 1 if t − s + 1 ≤ i ≤ t, bij = 0 otherwise. Then α + β has matrix (cij ) with cii = 1 if 1 ≤ i ≤ t − s cii = 2 if t − s + 1 ≤ i ≤ r, cii = 1 if r + 1 ≤ i ≤ t, cij = 0 otherwise. Thus rank α = r, rank β = s rank(α + β ) = t. If t ≤ r − 1, consider the linear map α given by the matrix (aij ) with aii = 1 if 1 ≤ i ≤ r, aij = 0 otherwise, and the linear map β given by the matrix (bij ) with bii = −1 if 1 ≤ i ≤ r − t, bii = 1 if r − t + 1 ≤ i ≤ s, bij = 0 otherwise. Then α + β has matrix (cij ) with cii = 2 if r − t + 1 ≤ i ≤ s cii = 1 if s + 1 ≤ i ≤ r, cij = 0 otherwise. Thus rank α = r, rank β = s rank(α + β ) = t. 116 Exercise 5.5.19 To avoid repetition, we go straight to (iii). We consider n×n matrices and say A ∈ Γ if and only if there is a K such that n=1 arj = K for r all j and n=1 air = K for all i. r (i)′ Observe that n A, B ∈ Γ ⇒ ⇒ n n arj = r =1 n air , r =1 n bij = i=1 n λarj + µbrj = r =1 r =1 j =1 bij ∀i, j λir + µbir ∀i, j λA + µB ∈ Γ. Since 0 ∈ Γ, we have γ a vector space. (ii)′ Let Γ0 be the subset of Γ such that n n arj = r =1 air = 0. r =1 We observe that Γ0 is a subspace of Γ since 0 ∈ Γ0 and n A, B ∈ Γ ⇒ n arj = r =1 n air = 0, r =1 bij = i=1 n ⇒ n j =1 bij = 0 ∀i, j 3 λarj + µbrj = r =1 r =1 λir + µbir = 0 ∀i, j λA + µB ∈ Γ0 . We note that if E is the n × n matrix all of whose entries are 1 then any basis of Γ0 together with E gives a basis for Γ. If 1 ≤ i, j ≤ n − 1, let E (i, j ) be the n × n matrix with entry 1 in the (i,j)th place and (n, n)th place and entry −1 in the (i, n)th place and (n, j )th place. We observe that E (i, j ) ∈ Γ0 . We observe by looking at the (r, s)th entry that 1≤i,j ≤n−1 aij E (i, j ) = 0 ⇒ ars = 0 ∀1 ≤ r, s ≤ n − 1, so the Ers are linearly independent. If A = (aij ) then A− aij E (i, j ) = 0 1≤i,j ≤n−1 117 We check the (n, n)th entry by observing that n n aij = 0 i=1 j =1 Thus the E (i, j ) form a basis for Γ0 and the Ei,j [1 ≤ i, j ≤ n − 1] together with E form a basis for Γ. Thus Γ has dimension (n − 1)2 +1 = n2 − 2n + 1 I cannot see any natural basis. 118 Exercise 5.6.2 We have P (r) ≡ b0 + b1 cr P (r) ≡ 2 + 2cr P (1) ≡ 2 + 2 × 2 ≡ 6 P (2) ≡ 2 + 2 × 4 ≡ 3 (all modulo 7). P (3) ≡ 2 + 2 × 5 ≡ 5 (P (1), c(1)) and (P (2), c(2)) yield 6 ≡ b0 + 2b1 3 ≡ b0 + 4b1 Subtracting the first equation from the second yields 3 ≡ −2b1 . Euclid’s algorithm (or inspection) tells us to multiply by 3 to recover b1 = 2. Substitution yields b0 = 2. (P (1), c(1)) and (P (3), c(3)) yield 6 ≡ b0 + 2b1 5 ≡ b0 + 5b1 Subtracting the first equation from the second yields −1 ≡ −3b1 . Euclid’s algorithm (or inspection) tells us to multiply by 5 to recover b1 = 2. Substitution yields b0 = 2. 119 Exercise 5.6.3 If cj = ck with j = k , then two people have the same secret. More people are needed to find the full secret. The Vandermonde determinant 1 cr(1) c2(1) r 1 cr(2) c2(2) r 1 cr(3) c2(3) det r . . . . . . . . . 1 cr(k) c2(k) r k −1 . . . cr(1) k −1 . . . cr(2) k −1 . . . cr(3) . ... . . k− . . . cr(k1 ) does not vanish (and so the system is soluble) if and only if the cr(j ) are distinct. If cj = 0 then P (j ) = b0 and j knows the secret. The proof that fewer cannot find the secret depended on k −1 cr(1) c2(1) . . . cr(1) r k −1 c2(2) . . . cr(2) cr(2) r k −1 2 ≡ cr(1) cr(2) . . . cr(k−1) cr(3) cr(3) . . . cr(3) (cr(i) −cr(j ) ) ≡ 0 det . . . . ... 1≤j<i≤k−1 . . . . . . . . k −1 2 cr(k−1) cr(k−1) . . . cr(k−1) and this needs the cr(j ) distinct and non-zero. 120 Exercise 5.6.4 P (r) ≡ b0 + b1 cr P (r) ≡ 1 + cr P (1) ≡ 1 + 1 × 1 ≡ 2 P (2) ≡ 1 + 1 × 4 ≡ 5 all modulo 6. The recovery equations are 2 ≡ b0 + b1 5 ≡ b0 + 4b1 There are two solutions (b0 , b1 )) = (1, 1) and (b0 , b1 )) = (5, 3). If p is not a prime, mn ≡ k may have more than one solution in n even if k ≡ 0. 121 Exercise 5.6.5 (i) The secret remains safe. The arguments continue to work if the cj are known. (ii) If the bs are known for r = 0 the holder of any pair (cr , P (r)) can compute b0 directly as b0 ≡ P (r) − bs c s . r s=1 122 Exercise 5.6.6 With k rings, they can say nothing with high probability since one of the rings may be a fake so they would have only k − 1 truth telling rings. With k + 1 rings, either every set of k rings gives the same answer in which case all the rings are true and the answer is correct or each set of k rings will give a different answer (one of which corresponding to the true k rings will be correct). All they can say is that one ring is fake but they do not know which one or what the Integer of Power is. With k + 2 rings, there are (k + 2)(k + 1)/2 different sets of k rings. Either each set tells the same story in which case all the rings are true and the answer is correct or k + 1 sets will tell the same story in which case they are the k + 1 sets containing k true rings and their answer is correct and k (k − 1)/2 sets giving different answers. The heroes can identify the k + 1 correct rings since they belong to the truth telling sets. One prime to rule them all One prime to bind them In the land of composites Where the factors are 123 Exercise 6.1.2 Observe that, writing A = (aij ), we have a1j n a2j α ej = aij ei = . . . . i=1 anj 124 Exercise 6.1.3 Let A = (aij ), B = (bij ) and let αβ have matrix C = (cij ). Then n cij ei = αβ ej = α(β ej ) i=1 n bkj ek =α k=1 n = n bkj αek = k=1 n n bkj k=1 ark er r =1 so, by the uniqueness of the basis expansion, n cij = aik bkj k=1 i.e. C = AB . n = ark bkj r =1 k=1 er 125 Exercise 6.1.5 bij fi = α(fj ) = α(prj er ) = prj αer = prj asr es = prj asr qis fi = qis asr prj fi . Thus bij = qis asr prj . 126 Exercise 6.1.7 (i) Theorem 6.1.5 shows that, if A and B represent the same map with respect to two bases, then A and B are similar. If A and B are similar then B = P −1 AP . Let α be the linear map with matrix A corresponding to a basis e1 , e2 , . . . , e n . n i=1 Let fj = pij ei . Since P has rank n, the fj form a basis and Theorem 6.1.5 tells us that α has matrix B with respect to this basis. (ii) Follows at once from Theorem 6.1.5 and part (i). (iii) Write A ∼ B if B = P −1 AP for some non-singular P . I −1 AI = IAI = A so A ∼ A. If A ∼ B then B = P −1 AP for some non-singular P and writing Q = P −1 we have Q non-singular and A = Q−1 BQ so B ∼ A. If A ∼ B , B ∼ C then B = P −1 AP , C = Q−1 CQ for some nonsingular P , Q. Then P Q is non-singular and C = (P Q)−1 A(P Q) so C ∼ A. (iv) If A represents α with respect to one basis then A represents α with respect to the same basis. Thus A ∼ A. If A ∼ B then there exist two bases with respect to which A and B represent the same map so B ∼ A. By (i), if there exist two bases with respect to which A and B represent the same map then given any basis E1 there exists a basis E2 so that A and B represent the same map with respect to E1 and E2 . Thus if A ∼ B , B ∼ C we can find bases Ej such that A and B represent the same map with respect to E1 and E2 , and b and C represent the same map with respect to E2 and E3 , so A and C represent the same map with respect to E1 and E3 so A ∼ B . 127 Exercise 6.1.8 We have n pij ej = (pi1 , pi2 , . . . , pin )T . fj = i=1 128 Exercise 6.1.9 Observe that e1 = f1 , e2 = f2 − f1 , e3 = f3 − f2 , 1 1 12 1 α(f1 ) = −1 2 1 0 = −1 = e1 − e2 = 2f1 − f2 , 0 0 13 0 2 1 12 1 −1 2 1 1 = 1 α(f2 ) = 1 0 13 0 = 2e1 + e2 + e3 = f1 + f3 , 1 12 4 1 α(f3 ) = −1 2 1 1 = 2 0 13 4 1 = 4e1 + 2e2 + 4e3 = 2f1 − 2f2 + 4f3 . The new matrix is 2 −1 0 B = 1 0 1 . 2 −2 4 129 Exercise 6.1.12 Without Theorem 6.1.10, our definition would depend on the choice of basis. 130 Exercise 6.2.3 (i) Observe that det t ab 10 − cd 01 = det t − a −b −c t − d = (t − a)(t − d) − (b)(−c) = t2 − (a + d) + (ad − bc) = t2 − (Tr A)t + det A. (ii) We have 3 det(tI − A) = ζ (σ ) σ ∈S3 i=1 (tδiσi − aiσi ) = ut3 + vt2 + ct + w. Looking at the coefficient of t3 , we have u = 1. Looking at the coefficient of t2 , we have v = −a11 − a22 − a33 . Setting t = 0 w = det(−A) = (−1)3 det A = − det A. 131 Exercise 6.2.4 Write ∂P for the degree of a polynomial P . Observe that ∂biσ(i) = 0 if σ (i) = i and ∂biσ(i) = 1 if σ (i) = i. Thus if σ = ι we know that σ (i) = i for at least two distinct values of i so n n ∂ i=1 biσ(i) ≤ i=1 ∂biσ(i) ≤ n − 2. It follows that n det(tI −A) = ζ (σ )biσ(i) (t) = σ ∈Sn n bii (t) i=1 ζ (σ )biσ(i) (t) = σ =ι i=1 (t−aii )+Q(t), where Q is a polynomial of degree at most n − 2. Thus the coefficient of tn−1 in det(tI − A) is the coefficient of tn−1 in − aii ), that is to say, − Tr A. n i=1 (t The constant term of a polynomial P is P (0) so the constant term in det(tI − A) is det((−1)A) = (−1)n det A. 132 Exercise 6.2.9 We show that any linear map α : R2n+1 → R3 has an eigenvector. It follows that there exists some line l through 0 with α(l) ⊆ l. Proof. Since det(tι − α) is a real polynomial of odd degree, the equation det(tι − α) = 0 has a real root λ say. We know that λ is an eigenvalue and so has an associated eigenvector u, say. Let l = {su : s ∈ R}. 133 Exercise 6.2.11 det(tι − ρθ ) = det t − cos θ sin θ − sin θ t − cos θ = (t − cos θ)2 + sin2 θ = t2 − 2 cos θ + 1. The equation t2 − 2 cos θ + 1 = 0 has a real root if and only if 4 cos2 θ ≥ 4, so if and only if cos θ = ±1, i.e. if and only if θ ≡ 0 mod π . If θ ≡ 0 mod 2π , then ρθ = ι and every non-zero vector is an eigenvector with eigenvalue 1. If θ ≡ π mod 2π , then ρθ = −ι and every non-zero vector is an eigenvector with eigenvalue −1. 134 Exercise 6.2.14 (i) PA (t) = det(tI + A) = (−1)n det(−tI − A) = (−1)n PA (−t). (ii) Let δ be as in Lemma 6.2.13. Let tn = δ/2n. 135 Exercise 6.2.15 χAB (t) = det(tI − AB ) = det(A(tA−1 − B )) = det A det(tA−1 − B ) = det(tA−1 − B ) det A = det((tA−1 − B )A) = det(tI − BA) = χBA (t). In general, we can find sn → 0 such that sn I + A is invertible. Thus if t is fixed χ(sn I +A)B (t) = χB (sn −A) (t). But χ(sn I +A)B (t) → χAB (t) and χB (sn I +A) (t) → χBA (t) so χAB (t) = χBA (t). Allowing t to vary, we see that χAB = χBA . Observe that − Tr C is the coefficient of tn−1 in χC (t). Thus χA = χB ⇒ Tr A = Tr B . By the previous result this gives Tr AB = Tr BA. (i) (False in general.) We have A= 10 , B= 00 then ABC = (AB )C = 11 , C= 00 10 10 01 01 = and 00 11 11 00 00 00 00 . = 11 11 00 Thus Tr(ABC ) = 1 = 0 = Tr(ACB ) and χABC = χACB . ACB = (AC )B = (ii) (True.) χABC = χA(BC ) = χ(BC )A = χBCA . 136 Exercise 6.2.16 Let e = (1, 1, 1, . . . , 1)T . Observe that A is magic with constant κ if and only if Ae = AT e = κe. In particular e is an eigenvector of A. If A is magic with constant κ and AB = BA = I , then e = BAe = κB e, so κ = 0 and B e = κ−1 e. Since B T AT = (AB )T = I T = I , the argument of the previous sentence gives B T e = κ−1 e. Thus B is magic with constant κ−1 . Suppose A is magic with constant κ. If A is invertible Adj A = (det A)A−1 so, by the previous paragraph with B = A−1 , Adj A is magic with constant κ−1 det A. Chose tn → 0 with A + tn I invertible. If we write Adj(A + tn I ) = (cij (n)), Adj A = (cij ) then cij (n) → cij as n → 0. Thus n 0= n cir (n) − r =1 n r =1 ckr and n=1 cir − r Thus Adj A is magic. r =1 n ckr (n) → r =1 n cir − for all i and k . Similarly ckr r =1 n r =1 cir = n r =1 crk . 137 Exercise 6.3.2 (i) If x1 e1 + x2 e2 = 0, then 0 = α(x1 e1 + x2 e2 ) = x1 αe1 + x2 αe2 = λ1 x1 e1 + λ2 x2 e2 Thus 0 = λ1 (x1 e1 + x2 e2 ) − (λ1 x1 e1 + λ2 x2 e2 ) so that = (λ1 − λ2 )x2 e2 , (λ1 − λ2 )x2 = 0 and x2 = 0. Similarly x1 = 0, so e1 and e2 are linearly independent. (ii) Observe that 0 = (α − λ2 ι)(0) = (α − λ2 ι)(x1 e1 + x2 e2 ) = (λ1 − λ2 )x2 e2 and proceed as before. (iii) Observe that 0 = (α − λ2 ι)(α − λ3 ι)0 = (α − λ2 ι)(α − λ3 ι)(x1 e1 + x2 e2 + x3 e3 ) = (α − λ2 ι) (λ1 − λ3 )x1 e1 + (λ2 − λ3 )x1 e2 so that = (λ1 − λ2 )(λ1 − λ3 )x1 e1 (λ1 − λ2 )(λ1 − λ3 )x1 = 0 and x1 = 0. Similarly x2 = x3 = 0 so e1 , e2 , e3 are linearly independent. 138 Exercise 6.4.2 (i) Observe that det(tι − β ) = det t −1 0t = t2 . Thus the characteristic equation of β only has 0 as a root and β only has zero as an eigenvalue. (ii) If x = (x, y )T then β x = 0x ⇔ (y, 0)T = (0, 0)T ⇔ y = 0 Thus the eigenvectors of β are the non-zero vectors of the form (x, 0)T and these do not span R2 . 139 Exercise 6.4.5 (i) We have Q(t) = det so Q(A) = t−λ 0 0 t−µ = (t − λ)(t − µ) = t2 − (λ + µ)t + λµ, λµ 0 (λ + µ)λ 0 λ2 0 + 2− 0 λµ 0 (λ + µ)µ 0µ = 00 . 00 Thus Q(α) = 0. (ii) We have Q(t) = det so Q(A) = t − λ −1 0 t−λ λ2 0 0 λ2 − = (t − λ)2 = t2 − 2λt + λ2 , 2λ2 0 0 2λ2 + λ2 0 0 λ2 = 00 00 Thus q (α) = 0. (iii) We know that there exists a basis with respect to which α has one of the two forms discussed, so the result of Example 6.4.4 is correct. 140 Exercise 6.4.7 Let P be as in Theorem 6.4.6. Since P is non-singular, det P = 0, so (working in C) we can find a κ with κ2 = det P . Set ν = κ−1 and Q = νP . Then det Q = ν 2 det P = 1 and Q−1 AQ = ν −1 νP −1 AP = P −1 AP. 141 Exercise 6.4.8 (i) Since d −λt (e x(t)) = −λe−λt x(t) + e−λt x(t) = e−λt (−λx(t) + x(t)) = 0, ˙ ˙ dt the mean value theorem tells us that e−λt x(t) = C and so x(t) = Ce−λt for some constant C . (ii) Since d −λt e x(t) − Kt = e−λt (−λx(t) + x(t) − Keλt ) = 0, ˙ dt the mean value theorem tells us that e−λt x(t) − Kt = C and so x(t) = (C + Kt)e−λt for some constant C . 142 Exercise 6.4.9 If x1 (t) = x(t) and x2 (t) = x(t) then x1 (t) = x2 (t) automatically and ˙ ˙ x2 (t) = −bx1 (t) − ax2 (t) ⇔ x = −ax − bx ⇔ x + ax + bx = 0 ˙ ¨ ˙ ¨ ˙ det(λI − A) = λ(λ + a) + b = λ2 + aλ + b 143 Exercise 6.5.5 By definition, 1 2 1 = 2 1 = 2 1 = 2 Q−1 Rθ Q = 1i i1 cos θ − sin θ sin θ cos θ 1 −i −i 1 cos θ + i sin θ − sin θ + i cos θ sin θ + i cos θ cos θ − i sin θ eiθ ieiθ ie−iθ e−iθ 2eiθ 0 0 2e−iθ 1 −i −i 1 = eiθ 0 . 0 e−iθ 1 −i −i 1 144 Exercise 6.5.2 Since A has distinct eigenvalues, it is diagonalisable and we can find a an invertible matrix B such that BAB −1 = D, where D is the diagonal matrix with j th diagonal entry λj . If we set y = B x (using column vectors) we obtain ˙ ˙ y = B x = BAx = BAB −1 y = Dy. Thus yj = λj yj and yj = cj eλj t for [1 ≤ j ≤ n] and so, writing ˙ F = B −1 , F = (fij ) n x1 = n f1j yj = j =1 f1j cj exp(iλj t). j =1 If we set µj = f1j cj , we obtain the required result. If A = D, then x = y and x1 = c1 exp(iλ1 t) and, with the notation of the question, we have µj = 0 for 2 ≤ j ≤ n. 145 Exercise 6.6.2 Dq is the diagonal matrix with diagonal entries the q th powers of the diagonal entries of D. (ii) A and D represent the same linear map α with respect to two bases E1 and E2 with basis change rule U = P V P −1 . Aq and Dq represent the same linear map αq with respect to the two bases E1 and E2 so Aq = P Dq P −1 . 146 Exercise 6.6.7 (i) We have λ−q αq (x1 e1 + x2 e2 ) = x1 e1 + x2 (µ/λ)n x2 e2 → x1 e1 coordinate wise. (i)′ We have λ−q αq (x1 e1 + x2 e2 ) = x1 e1 + x2 (µ/λ)n x2 e2 , but (µ/λ)n does not tend to a limit, so λ−q αq (x1 e1 + x2 e2 ) converges coordinatewise if and only if x2 = 0. (ii)′ Nothing to prove. (iii) We have α (λq x1 + qλq−1 x2 )e1 + λq x2 e2 = λ(λq x1 + qλq−1 x2 )e1 + λq (x2 e1 + λe2 ) = λq+1 x1 + (q + 1)λq x2 e1 + λq+1 x2 e2 so, by induction αq (x1 e1 + x2 e2 ) = (λq x1 + qλq−1 x2 )e1 + λq x2 e2 for all q ≥ 0. Thus q −1 λ−q αq (x1 e1 + x2 e2 ) = (q −1 x1 + λ−1 x2 )e1 + q −1 x2 e2 → λ−1 x2 )e1 coordinatewise. (iv) Immediate. 147 Exercise 6.6.8 F1 = F2 = 1, F3 = 2, F4 = 3, F5 = 5, F6 = 8, F7 = 13, F8 = 21, F9 = 34, F10 = 55. F0 + 1 = F0 + F1 = F2 = 1, so F0 = 0. F−1 + 0 = F−1 + F0 = F1 = 1, so F−1 = 1. Write G−n = (−1)n+1 Fn . Then G−n+1 + G−n = (−1)n (Fn−1 − Fn ) = (−1)n−1 Fn−2 = G−n+2 and G−1 = F−1 , G(0) = F (0) so, by induction, F−n = G−n for all n ≥ 0. Thus F−n = (−1)n+1 Fn for all n. 148 Exercise 6.6.9 If b = 0, but a = 0, then un = (−a)n u0 . If a = b = 0, then un = 0 for all n. 149 Exercise 6.6.10 We look at vectors u(n) = un . un+1 We then have u(n + 1) = Au(n), where A = and so 0 1 −a −b u0 un . = An u1 un+1 Now A is not a multiple of the identity matrix so (since the roots are equal) there is a basis e1 e2 with Ae1 = λe1 = λe1 and Ae2 = λe1 + λe2 . Now An (x1 e1 + x2 e1 ) = (λn x1 + nλn−1 x2 ) + λn x2 e2 so, since u(n) = An u(0), we have un = (c + c′ n)λn for some constants c and c′ depending on u0 and u1 . 150 Exercise 6.6.11 We look at vectors u(n) = (un , un+1 , . . . , un+m−1 )T . We then have 0 0 . . . u(n + 1) = Au(n), where A = 1 0 . . . 0... 1... . . . 0 0 ... . . . 0 0 1... 1 −am−1 −am−2 −am−3 . . . −a0 The characteristic polynomial of A is P (t), so, since all the roots of P are distinct and non-zero, the eigenvectors ej (corresponding to an eigenvalue λj ) of A form a basis. Setting u0 = m k=1 bk ek we get m n un = A u0 = bk λ k e k k=1 and so looking at the first entry in the vectors m c k λn k un = k=1 for some constants cj . Conversely, if m c k λn , k un = k=1 then m−1 ur + m n λk −m P (λk ) = 0. aj uj −m+r = j =0 k=1 Observe that, if |λ1 | > |λk | for 2 ≤ k ≤ m, then m λ − n un = 1 k=1 ck (λ−1 λk )n → c1 1 as n → ∞. Thus, if c0 = 0, we have ur−1 = 0 for r large and ur → λ1 ur − 1 as r → ∞. A very similar proof shows that, if |λk | > |λm | for 1 ≤ k ≤ m − 1 and cm = 0, then un = 0 for −n large and ur − 1 → λm . ur 151 Exercise 6.6.12 (i) Observe that √ √ √ ( 5)2 − 12 1 + 5 −1 + 5 × = = 1. 2 2 4 (ii) We have det(tI − A) = t(t − 1) − 1 = t2 − t − 1 = (t − τ )(t + τ −1 ), so the eigenvalues are τ and −τ −1 . If (x, y )T is an eigenvector with eigenvalue τ , y = τx x + y = τ y, so e1 = (1, τ )T is an eigenvector with eigenvalue τ . A similar argument shows that e2 = (−1, τ −1 )T is an eigenvector with eigenvalue τ −1 . By inspection (or solving two simultaneous linear equations) 0 1 1 = √ (e1 + e2 ). 5 (iii) It follows that Fn Fn+1 = An F0 F1 = 5−1/2 (τ n e1 + (−τ )−n e2 ) and so Fn = 5−1/2 (τ n + (−τ )−n ) (vi) Since Fn is an integer and |5−1/2 (−τ )−n | ≤ 5−1/2 < 1/2 for n ≥ 0 we know that Fn is the closest integer to τ n /51/2 . Both methods give F20 = 6765. (v) Observe that An has first column An−1 (F0 , F1 )T = (Fn−1 , Fn )T and second column An−1 (F1 , F2 )T = (Fn , Fn+1 )T . (Or use induction.) 2 (−1)n = (det A)n = det An = Fn−1 Fn+1 − Fn . 152 (vi)True for all n because Fn = (−1)n+1 F−n . (vi) We have A2n = An An , so Fn−1 Fn Fn Fn+1 Fn−1 Fn Fn Fn+1 = F2n−1 F2n F2n F2n+1 so, evaluating the upper left hand term of the matrix, we get 2 2 Fn + Fn−1 = F2n−1 . 153 Exercise 6.6.13 (n) (i) Let eij be the number of routes from i to j involving exactly n (n) (n) flights. We claim that eij = dij . We use induction on n. The result is clearly true if n = 1. Suppose it is true for n = r. (r +1) eij = number of journeys with exactly r + 1 flights from i to j m = number of journeys with exactly r + 1 flights k=1 from i to j ending with a flight from k to j = number of journeys with exactly r flights from i to k dkj =0 m (r ) = (r +1) dik dkj = dij k=1 The result follows. In particular, there is a journey from i to j which involve exactly n (n) flights if and only if dij > 0. (ii) We can produce an argument like (i) if we make the convention that a flight includes just staying at an airport (so a flight from i to i is allowed). In particular there is a journey from i to j which involve n flights or ˜(n) less (between different airports) if and only if dij > 0. (iii) We write down the safe states for example i = (G|P W C ) goat on one side, peasant, wolf and cabbage on other and write dij = 1 if the peasant can change the situation from i to j in one crossing. Otherwise dij = 0. The least number of crossings is the least N for which dN W C |∅),(∅|GP W C ) = 0. (GP Observe that, since the peasant will not repeat a state, if there are M safe states the problem is soluble, if at all, with at most M − 1 crossings so the method will reveal if the problem is insoluble. (The cabbage survived but was so distressed by its narrow escape that it turned over a new leaf). 154 Exercise 6.7.1 1≤i≤n (i) Suppose that A = (aij )1≤i≤n is lower triangular. Then, by row expansion, det A = a11 det B 1≤i≤n−1 where B = (ai+1,j +1 )1≤i≤n−1 . Since B is lower triangular with diagonal entries ai+1,i+1 [1 ≤ i ≤ n − 1] the required result follows by induction. (ii) We have n L invertible det L = 0 ⇔ j =1 ljj = 0 ⇔ ljj = 0 ∀j (iii) Since tI − L is lower triangular n det(tI − L) = j =1 (t − ljj ) so the roots of the characteristic equation are the diagonal entries ljj (multiple roots occurring with the correct multiplicity). Observe that (0, 0, 0, . . . , 0, 1)T is an eigenvector with eigenvalue lnn . 155 Exercise 6.7.2 Just repeat the earlier discussion mutatis mutandis. The following is merely a variation. If the system is n uij xj = yi j =1 then setting gular system x′j = xn−j , ′ yi = yn−i lij = un−i,n−j gives the lower triann ′ lij x′j = yi . j =1 156 Exercise 6.7.3 (i) Observe first that the product of two lower triangular matrices is triangular since if aij , bij = 0 for i + 1 ≥ j , then n air brj = air brj = 0 j ≤r ≤i r =1 for i + 1 ≥ j . We use induction to show that the inverse of an invertible n × n lower triangular matrix is triangular. Since every 1 × 1 matrix is lower triangular, the result is certainly true for n = 1. Suppose it is true for n = m and A is an m + 1 × m + 1 invertible lower triangular matrix. Since A is invertible a11 = 0. If B = (bij ) is the m + 1 × m + 1 matrix with b11 = a−1 , bi1 = −ai1 a−1 11 11 bii = 1 for 2 ≤ i ≤ m + 1 and bij = 0, otherwise, then BA = 1 0T 0L where L is an m × m invertible lower triangular matrix and 0 ∈ Rm is the zero column vector. If we now set 1 0T 0 L− 1 is lower triangular (by the inductive hypothesis), we is lower triangular so C −1 B is lower triangular. But C= then since L−1 know that C −1 (C −1 B )A = C −1 (BA) = C −1 C = I so A has lower triangular inverse and the induction is complete. (ii) By the first paragraph of (i), A, B ∈ L ⇒ AB ∈ L. By inspection I ∈ L and, by part (i), A ∈ L ⇒ A−1 ∈ L. 157 Exercise 6.7.6 Observe that 1 2 LU = −1 2 4 = −2 2 4 = −2 00 21 1 1 0 0 −1 −2 −3 1 0 0 −4 1 1 2−1 2−2 −1 + 3 −1 + 6 − 4 11 1 0 . 21 158 Exercise 6.7.7 Consider an n × n matrix A. The first step will be impossible only if the entire first row consists of zeros in which case it is clear that the n × n matrix A is not invertible. If the first step can be done, then we repeat the process with a n − 1 × n − 1 square matrix B and the same remarks apply to B . Thus either the process will stop because the matrix with which we deal at the given stage has top row all zeros or the process will not stop in which case we obtain an upper triangular matrix U and a lower triangular matrix L both with all diagonal entries non-zero so invertible. Since the product of invertible matrices is invertible A = LU is invertible. Thus the process works if and only if A is invertible. 159 Exercise 6.7.9 Suppose that a0 cd Then uv w0 = 01 . 10 01 au av . = 10 cu cv + dw It follows that au = 0 and av = 1, so u = 0, which is incompatible with cu = 1. Our initial formula cannot hold. 160 Exercise 6.7.10 We use induction on n. The result is trivial for n = 1 since (1)(u1 ) = (1)(u2 ) ⇒ (u1 ) = (u2 ) ⇒ u1 = u2 . Now suppose the result is true for n = m and L1 , L2 are lower triangular (with diagonal entries 1) and U1 , U2 upper triangular m + 1 × m + 1 matrices with L1 U1 = L2 U2 . Observe that we can write u(j ) uT 1 0T j Lj = and Lj = ˜j ˜j lj L 0 U ˜ ˜ where the Lj , L2 lower triangular (with diagonal entries 1) and the Uj m upper triangular m × m matrices, u(j ) ∈ F and lj , uj ∈ F are column vectors. Equating first rows in the equation L1 U1 = L2 U2 we get (u(1), uT ) = (u(2), uT ) 1 2 and equating first columns we get (1, lT )T = (1, lT )T 1 2 so u(1) = u(2), u1 = u2 , l1 = l2 . The condition L1 U1 = L2 U2 now gives ˜˜ ˜˜ L1 U1 = L2 U2 so by the inductive hypothesis ˜ ˜˜ ˜ L1 = L2 , U1 = U2 . We have shown that L1 = L2 , U1 = U2 and this completes the induction 161 Exercise 6.7.11 Since det LU = det L det U and det L, det U are the product of the diagonal entries, det LU is cheap to calculate. By solving the n sets of simultaneous linear equations n lir xrj = δij r =1 each of which requires of the order of n2 operations we can compute X = L−1 in the order of n3 operations and we can compute U −1 similarly. We now compute (LU )−1 = U −1 L−1 again in the order of n3 operations. 162 Exercise 6.7.12 (i) It is probably most instructive to do this ab initio, but quicker to use known results. ˜ We know that (after reordering columns) A = LU with L lower ˜ triangular and all diagonal entries 1 and U upper triangular. Since A ˜ is non-singular U is, so all its diagonal entries are non-zero. Let D be ˜ the diagonal matrix with diagonal entries the diagonal entries of U . −1 ˜ Set U = D U . Then U is upper triangular with all diagonal entries 1, D is non-singular and A = LDU. Suppose that Lj is a lower triangular n × n matrix with all diagonal entries 1, Uj is an upper triangular n × n matrix with all diagonal entries 1 and Dj is a non-singular diagonal n × n matrix [j = 1, 2]. We claim that, if L1 D1 U1 = L2 D2 U2 , then L1 = L2 , U1 = U2 and D1 = D2 . To see this observe that Dj Uj is non singular upper triangular so, by Exercise 6.7.10, L1 = L2 and D1 U1 = D2 U2 . By looking at the diagonal entries of both sides of the equation we get D1 = D2 so − − U1 = D1 1 (D1 U1 ) = D2 1 (D2 U2 ) = U2 . (ii) Yes. Either repeat the argument mutatis mutandis or argue as follows. ˜ ˜ If G = (gij ) is an n × n matrix, write G = (gn+1−i,n+1−j ). If G = LU ˜˜ ˜ with L lower triangular, U upper triangular, then G = LU and L is ˜ upper triangular whilst U is lower triangular. (iii) No. Observe that ab c0 x0 yz = ax + by bz . cx 0 163 Exercise 7.1.4 (i) Any set of orthonormal vectors is linearly independent. Any set of k linearly independent vectors forms a basis for U . (ii) By (i), we can find λj such that k x= λj ej . j =1 Now observe that k x, ei = λj ej , ei = λi . j =1 164 Exercise 7.1.6 If we take e1 = 3−1/2 (1, 1, 1), e1 = 2−1/2 (1, −1, 0), then, by inspection, they form an orthonormal system. We now use Gramm-Schmidt for x = (1, 0, 0). x − x, e1 e1 − x, e2 e2 = (1, 0, 0) − 3−1 (1, 1, 1) − 2−1 (1, −1, 0) say, Setting = (1/6, 1/6, −1/3) = a, e3 = a −1 a = 6−1/2 (1, 1, −2) we have e1 , e2 , e3 orthonormal. It follows at one that e1 = 3−1/2 (1, 1, 1), 2−1/2 (0, −1, 1), 6−1/2 (−2, 1, 1) is another orthonormal system. 165 Exercise 7.1.8 (Dimension 2) If a point B does not lie in a plane π , then there exists a unique line l′ perpendicular to π passing through B . The point of intersection of l with π is the closest point in π to B . More briefly, the foot of the perpendicular dropped from B to π is the closest point in π to B . (Dimension 1) If a point B does not lie on a line l, then there exists a unique line l′ perpendicular to l passing through B . The point of intersection of l with l′ is the closest point in l to B . More briefly, the foot of the perpendicular dropped from B to l is the closest point in l to B . 166 Exercise 7.1.9 (i) We have 2 k x− λj ej j =1 k x− k λj ej , x − j =1 λj ej j =1 k 2 =x −2 k j =1 j =1 k 2 =x − k x, ej 2 + j =1 j =1 k 2 ≥x − x, ej 2 j =1 with equality if and only if λj = x, ej . (ii) It follows from (i) that k 2 x ≥ x, ej 2 j =1 with equality if and only if λj = x, ej and k x− λj ej = 0, j =1 ie λ2 j λj x, ej + k x= λj ej j =1 and this occurs if and only if x ∈ span{e1 , e2 , . . . , ek }. (λj − x, ej )2 167 Exercise 7.1.10 Observe first that 0, u = 0 ⊥ for all u so 0 ∈ U . Next observe that if λ1 , λ2 ∈ R v1 , v2 ∈ U ⊥ ⇒ λ1 v1 + λ2 v2 u = λ1 v1 , u + λ2 v2 , u = 0 + 0 = 0 for all u ∈ U ⇒ λ1 v1 + λ2 v2 ∈ U ⊥ . If we take a − b = v in Theorem 7.1.7, we see that a ∈ Rn can be written in one and only one way as x = u + v with u ∈ U , v ∈ U ⊥ . If e1 , . . . , ek is a basis for U and ek+1 , . . . , el is a basis for U ⊥ , the previous paragraph tells us that e1 , . . . , ek , ek+1 , . . . , el is a basis for Rn so l = n and dim U + dim U ⊥ = n. 168 Exercise 7.2.2 n i=1 n j =1 j =1 n n n i=1 n αx, y = aij xj yi cji xj yi = j =1 i=1 = aij xj yi = x, α∗ y . 169 Exercise 7.2.6 Suppose that A, B represent α, β with respect to some basis. Lemma 7.2.4 enables us to translate Lemma 7.2.5 into matrix form (i) Since (αβ )∗ = β ∗ α∗ , we have (AB )T = B T AT . (ii) Since α∗∗ = (α∗ )∗ = α, we have AT T = (AT )T + A. (iii) Since (λα +µβ )∗ = λα∗ +µβ ∗ , we have (λA+µB )T = λAT +µB T . (vi) Since ι∗ = ι, we have I T = I . In coordinates (i) If AB = C , then n cij = air brj r =1 so n cji = brj air r =1 and (AB )T = C T = B T AT . (ii) AT T = (aji )T = (aij ) = A. (iii) (λA + µB )T = (λaji + µbji ) = λ(aji + µbji ) = λAT + µB T . (iv) δij = δji so I = I T . 170 Exercise 7.2.10 Let the j th row of A be aj . A ∈ O(Rn ) ⇔ AAT = I ⇔ ai aT = δij ⇔ ai · aj = δij ⇔ the aj are o.n. j Thus (i)⇔(ii). Since A ∈ O(Rn ) ⇒ AT T AT = AAT = I ⇒ AT ∈ O(Rn ) transposition gives (i)⇔(iii). We turn to the statements. (i) FALSE. Consider 1 −1 . 22 (ii) FALSE. Consider 12 . 12 (iii) FALSE If a1 , a2 , . . . , an−1 , an are the rows of an orthogonal matrix, so are a1 , a2 , . . . , an−1 , −an (and an = 0). (iv) TRUE If a1 , a2 , . . . , an−1 are orthonormal in Rn then there are exactly two choices ±an giving an orthonormal set and of these exactly one will give det A = 1. 171 Exercise 7.2.13 Take 11 11 . , B= 0 −1 01 Note that the columns in each matrix are not orthogonal. A= 172 Exercise 7.3.2 Any pair of orthonormal vectors form a basis for R2 . We observe that e1 , e1 = 1, e1 , −e2 = − e1 , e2 = 0, −e2 , −e2 = e2 , e2 = 1. Observe that αe1 = cos θe1 + sin θe2 = cos θe1 − sin θ(−e2 ) = α(−e2 ) = −αe2 = sin θe1 − cos θe2 = sin θe1 + cos θ(−e2 ) so, with respect to the new basis, α has matrix cos θ sin θ − sin θ cos θ = cos(−θ) − sin(−θ) sin(−θ) cos(−θ) 173 Exercise 7.3.4 (i) We have cos θ − sin θ sin θ cos θ x y = x cos θ − y sin θ x sin θ + y cos θ which I was told in school was a rotation through θ We have −x x −1 0 = y y 01 which certainly looks like a reflection in the y axis to me. Since we may take e1 , u1 e1 , u1 2 + e1 , u2 2 = 1, = cos φ, e1 , u2 = sin φ. We have e1 = cos φu1 + sin φu2 so e1 = −αe1 = − cos φαu1 − sin φαu2 = − cos φ(cos θu1 + sin θu2 ) − sin φ(sin θu1 + cos θu2 ) Taking the inner product with u1 we get so cos φ = − cos φ cos θ − sin φ sin θ = − cos(θ − φ) and θ = 2φ − π . φ=π+θ−φ 174 Exercise 7.3.6 Since a reflection in a plane has perpendicular eigenvectors with eigenvalues −1, 1, 1, α is a reflection only if det α = −1 so, by our discussion, α has matrix −1 0 0 A = 0 cos θ − sin θ , 0 sin θ cos θ so t+1 0 0 t − cos θ sin θ = (t−1)(t2 −2t cos θ +1) det(tι−α) = det 0 0 − sin θ t − cos θ and we can only have the correct eigenvalues if cos θ = 1. In this case we clearly have a reflection in the plane spanned by e2 and e3 . By definition α is a reflection in the origin if and only if α = −ι so if and only if cos θ = −1. 175 Exercise 7.3.7 det(tι − α) = det t − cos φ sin φ t − cos θ sin θ det − sin φ t − cos φ − sin θ t − cos θ = (t2 − 2t cos θ + 1)(t2 − 2t cos φ + 1) with no real roots unless cos θ = cos φ = 1 i.e. θ ≡ φ ≡ 0 (mod π ). 176 Exercise 7.4.3 Let u = x, n n and v = x − u. Then v, n = x − u, n = x − x, n n, n = x, n − x, n n, n = 0 as required. We observe that ρ(x) = u + v − 2( u, n + v, n) n = u + v − 2 x, n n = v − u. 177 Exercise 7.4.6 Let A be the point given by the vector a and B be the point given by the vector b. Let OM be an angle bisector of the line OA, OB of length 1. and let OM be given by the vector m. Let n be a unit vector perpendicular to m. 178 Exercise 7.4.9 The product of two reflections in lines OA, OB through the origin making an angle ∠AOB = θ is a rotation through 2θ. To prove this, observe that, if ρ1 and ρ2 are reflections, then ρ2 ρ1 preserves lengths and has determinant 1, so is a rotation. Now let e1 and e2 be orthonormal and let n = −e1 , m = − cos θe1 − sin θe2 . If ρ1 (x) = x − 2 x, n n then ρ2 (x) = x − 2 x, m m, ρ2 ρ1 e1 = ρ2 (e1 ) = e1 − 2 e1 , m m = e1 − 2 cos θ(cos θ1 + sin θ2 ) = cos 2θ1 + sin 2θ2 . Thus the rotations through angle θ are obtained from two reflections in lines at angle θ/2 (in the correct sense). 179 Exercise 7.5.1 (i) Observe that n Ax − b 2 = j =1 (x − bj )2 n n 2 = nx − 2x j =1 j =1 2 n =n x−n b2 j bj + −1 bj n + j =1 j =1 is minimised by taking x = n 2 n b2 j −n n −1 bj j =1 n j =1 bj . −1 In the case of the mountain this amounts to the reasonable choice of the average of the measured heights. (ii) Observe that Ax − b with 2 = f (x1 , x2 ) n f (x1 , x2 ) = i=1 (x1 − vi x2 − bi )2 We observe that f (x1 , x2 ) → ∞ as |x1 |, |x2 | → ∞. Since ∂f (x1 , x2 ) = ∂x1 n i=1 n 2(x1 − vi x2 − bi ) = 2 nx1 − bi i=1 and ∂f (x1 , x2 ) = − ∂x2 n i=1 n 2vi (x1 − vi x2 − bi ) = −2 x2 n 2 vi i=1 − bi v i , i=1 we see that f and therefore Ax − b has a unique minimum when (x1 , x2 ) = (µ, κ). We have sought to find a line u + x1 v = x2 ‘close to’ the points (ui , vi ) by minimising n i=1 (ui − x1 − x2 vi )2 . 180 Exercise 7.5.2 Observe that n f (x) i=1 2 m j =1 aij xj − bi is a smooth function of x with f (x) → ∞ as x → ∞. Differentiating and solving the resulting n linear equations in n unknowns will, in general (to see possible problems, take aij = 0 for all i and j ) give a unique stationary point which will be a minimum. The other two penalty functions cannot (in general) be minimised by calculus techniques, but could be attacked as linear programming problems. 181 Exercise 7.5.3 (i) If we write the columns of A as column vectors a1 , a2 , . . . am , then (possibly after reordering the columns of A), the Gram–Schmidt method gives us orthonormal row vectors e1 , e2 , . . . , ek such that a1 = r11 e1 a2 = r12 e1 + r22 e2 a3 = r13 e1 + r23 e2 + r33 e3 . . . ak = r1k e1 + r2k e2 + r3k e3 + . . . + rkk ekk for some rij [1 ≤ j ≤ i ≤ k ] with rii = 0 for 1 ≤ i ≤ k . and ap = r1p e1 + r2p e2 + r3p e3 + . . . + rpk ek for rij [1 ≤ i ≤ k, k + 1 ≤ p ≤ m]. We set rij = 0 in all the cases with 1 ≤ i ≤ n and 1 ≤ j ≤ m where this has not previously been defined. Using the Gram–Schmidt method again, we can now find ek+1 , ek+2 , . . . , en so that the vectors ej with 1 ≤ j ≤ n form an orthonormal basis for the space Rn of column vectors. If we take Q to be the n × n matrix with j th column ej , then Q is orthogonal and condition ⋆ gives A = QR. (ii) We have rii = 0 ∀ 1 ≤ i ≤ m ⇔ rank R = m ⇔ rank QT A = m ⇔ rank A = m since the linear map corresponding to QT is an isomorphism. 182 Exercise 7.5.4 (i) We know that π = {At : t ∈ Rn } is a subspace of Rm . Thus (e.g. by Theorem 7.1.7) it contains a closest point y to b. Choose x0 with Ax0 = y. to obtain a minimiser for our problem. Since A has rank strictly less than m we can find a non zero u ∈ Rm with Au = 0. Then every choice x = x0 + λz minimises Ax − b . (ii) The QL result is just the QR result with a renumbering of the basis. If A is an m × n matrix, Q an n × n orthogonal matrix and R an n × m upper triangular matrix with AT = QR, then A = RT QT , QT is orthogonal and RT a lower triangular matrix. Thus the QR results imply the corresponding LQ results and the QL results imply the corresponding RQ results. We could have used the QL in the discussion. However, the RQ and LQ results give Q in the wrong place when we derive the opening formulae. In addition (at least if we follow the obvious path) we factorise an n × m matrix with n ≥ m and we are not usually interested in cases where we have more unknowns than equations. 183 Exercise 7.5.5 Since ρ is norm preserving, ρ(a) = b ⇒ a = ρ(a) = b . Now take a = b with b = ±a and set c = (a − b)/2. Suppose that ρx = λx + µ c, x c describes a reflection with ρa = b and ρx = x whenever x, a = x, b . If x ⊥ c then x = ρ(x) = λx so λ = 1. Since 1 c, a = 2 ( a we have ρ(c) = c, so 2 + a, b ) = 1 ( b 2 2 + b, a ), c = c + µ c, c c 2 and µ = −2 c . Thus ρx = x − 2 c, x c c2 ρx = x − 2 c, x c, c2 Conversely, if then ρc = −c and d ⊥ c ⇒ ρd = d, so ρ is a reflection. Further x, a = − x, b ⇔ x, c = 0 ⇔ ρx = −x and c, a c c2 c, a + c, b c =a− c2 c, c =a−2 c c2 = a − 2c = b ρa = a − 2 and, since ρ is a reflection ρb = ρa. We have so Tij xj = xi − 2cj xj c −2 ci = (δij − 2(ck ck )−2 ci cj )xj , Tij = δij − 2(ck ck )−2 ci cj . 184 If a = (± a , 0, 0, . . . , 0), take T1 = I . Otherwise, take T1 = T with T as above. If C is an n × m matrix with cij = 0 for all j > i when i ≤ r, consider the (n − r) × (m − r) matrix A formed by the cij with n ≥ i ≥ n − r + 1 m ≥ j ≥ m − r + 1. We can find a (n − r) × (n − r) reflection matrix (or the identity) T with T A having first column zero except possibly the first entry. If S is the n × n matrix given by S= I0 , 0A then H = SC is an n × m matrix with hij = 0 for all j > i when i ≤ r + 1. Thus we can find reflection (or identity) matrices Tj such that Tn−1 Tn−2 . . . T1 A = R is upper triangular. The matrix Q = T1 T2 . . . Tn−1 is the product of reflection (or identity) so of orthonormal matrices so (since O(Rn ) is a group under multiplication) an orthonormal matrix. We have − − − A = (Tn−1 Tn−2 . . . T1 )−1 R = T1 1 T2 1 . . . Tn 1 R = T1 T2 . . . Tn R = QR. 185 Exercise 7.5.6 We are interested in the vector a = (1, 2, 2)T of length 3 which we wish to reflect to the vector b = (3, 0, 0)T . We thus want a reflection in the plane perpendicular to c = (a − b)/2 = (−1, 1, 1)T . This reflection is given by with matrix ρ(x) = x − 2 c −2 (c · x)c T = I − 2 c 2 cT c 100 1 −1 −1 2 1 = 0 1 0 − −1 1 3 −1 1 001 1 1 2 2 3 = 2 3 2 3 Now 3 1 3 2 −3 3 −2 3 1 3 4 3 11 3 3 T A = 0 0 1 0 1 −1 3 so (interchanging the second and third row) T A is upper triangular. 186 Exercise 7.5.7 We are interested in the vector a = (2, 2, −1)T of length 3 which we wish to reflect to the vector b = (3, 0, 0)T . We thus want a reflection in the plane perpendicular to 1 c = (a − b)/2 = (− 2 , 1, − 1 )T . 2 Setting d = 2c = (−1, 2, −1)T , reflection is given by ρ(y) = y − 2 d 2 d · yd with matrix Setting T = I − 2 c 2 dT d 100 1 −2 1 1 = 0 1 0 − −2 4 −2 3 001 1 −2 1 2 2 1 −3 3 3 1 = 2 −3 2 3 3 1 1 −3 2 −3 3 10 0 0 2 1 0 2 −3 3 3 Q= 2 1 2 . 0 −3 3 3 1 1 0 −3 2 −3 3 We see that (since Q is orthogonal) we seek the least squares fit for 4 13 1 0 2 Q 0 2 x = Q 4 1 0 −1 that is to say 1 0 0 0 so we require x2 = 1, x1 = 1. 4 3 3 3 x= 0 0 3 0 Let f be the square of the norm of 13 4 0 2 1 0 2 x − 4 0 −1 1 187 Then f (x1 , x2 ) = (x1 + 3x2 − 4)2 + (2x2 − 1)2 + (2x2 − 4)2 + (x2 + 1)2 and the unique stationary value of f (which must be minimum) is given by ∂f = 2(x1 + 3x2 − 4) 0= ∂x1 ∂f 0= = 2 3(x1 + 3x2 − 4) + 2(2x2 − 1) + 2(2x2 − 4) + (x2 + 1) ∂x2 that is to say, by 4 = x1 + 3x2 21 = 3x1 + 18x2 , that is to say, by 4 = x1 + 3x2 7 = x1 + 6x2 , so we require x2 = 1, x1 = 1, as stated earlier. 188 Exercise 8.1.7 (i) We have and AT = A ⇒ (P T AP )T = P T AT P T T = P T AP (P T AP )T = P T AP ⇒ P T AT P = P T AP ⇒ P (P T AT P )P T = P ((P T AP )P T ⇒ AT = A. (ii) If P ∈ SO(Rn ), then set Q = P . If P ∈ SO(Rn ), then set Q = P D. We have / det Q = det P det D = (−1)2 = 1 and QT Q = DT P T P D = DID = I , so Q ∈ SO(Rn ). Further, since P T AP is diagonal, QT AQ = DT P T AP D = D(P T AP )D = P T AP and QT AQ is diagonal. 189 Exercise 8.2.2 n j =1 If f akj = ajk for all j and k , for all k , but λ = µ, then n λ akj uj = λuk and n uk v k = k=1 λuk vk k=1 n n n = uj ajk vk = j =1 k=1 n =µ µuj vj j =1 n uj v j = µ j =1 uk vk = 0. ajk uj vk k=1 j =1 n k=1 j =1 n n n k=1 n akj uj vk = = so n j =1 uk v k , k=1 akj vj = µvk 190 Exercise 8.2.6 We have (by Cramer’s rule or direct calculation), P −1 = so 1 −1 01 10 11 Observe that P is not orthogonal. P AP −1 = P (AP −1 ) = 2 −2 01 = 2 −2 . 2 −1 191 Exercise 8.2.7 Let A= so AT = A. 1i i −1 det(tI − A) = (t − 1)(t + 1) + 1 = t2 so all the eigenvalues of A are 0 and, if A were diagonalisable, we would have A = 0 which is not the case. 192 Exercise 8.2.9 Interchanging x2 and x3 coordinates is the same as reflecting in a plane passing through the x1 axis and a line bisecting the angle between the x2 and x3 axes (so perpendicular to (0, 1, −1)T ). 193 Exercise 8.2.10 If there are n distinct eigenvalues λj with associated eigenvectors ej then P T AP = D with P orthogonal and D diagonal if and only if the columns of P consist of the eigenvectors ±ej (with choice of signs) in some order so there are exactly 2n n! such matrices. If A has less than n distinct eigenvalues then Rn has a basis of eigenvectors ej with en−1 and en having the same eigenvalues, then, if P has rth column er for 1 ≤ r ≤ n − 2 n − 1th column cos θen−1 − sin θen and nth column sin θen−1 + cos θen , we have P orthogonal and P T AP = D so case (ii) occurs. 194 Exercise 8.3.2 Observe that there exists a diagonal matrix D and an orthogonal matrix P with λ1 0 D= and A = P T AP 0 λ2 where λ1 , λ2 are the eigenvalues of A. Parts (i), (ii) and (iii) can be read off from the observation that det A = det D = λ1 λ2 and Tr A = Tr D = λ1 + λ2 . (iv) det A = uw − v 2 ≤ uw, so det A > 0 ⇒ u = 0. If u > 0 and det A > 0, then w > 0 so Tr A > 0, so, by (iii), the eigenvalues of A are strictly positive. If u < 0 and det A > 0, then w < 0, so Tr A < 0, so, by applying (iii) to A, the eigenvalues of A are strictly positive. Exercise 8.3.3⋆ 195 Exercise 8.3.4 By an orthogonal transformation we may reduce the equation ax2 + 2bxy + cy 2 = d to Ax2 + By 2 = K . By multiplying by −1 if necessary we may assume K ≥ 0 and by interchanging x and y if necessary we may assume that A = 0 ⇒ B = 0 and if A and B are non-zero with opposite signs then A > 0 > B The cases may now be enumerated as follows. (1) K > 0, A > 0, B > 0 ellipse. (2) K > 0, A < 0, B < 0 empty set. (3) K > 0, A > 0, B < 0 hyperbola. (4) K = 0, A > 0, B > 0 single point {0}. (5) K = 0, A < 0, B < 0 single point {0}. (6) K = 0, A > 0, B < 0 pair of lines meeting at 0. (7) K > 0, A > 0 B = 0 pair of parallel lines. (8) K > 0, A < 0, B = 0 empty set. (7) K = 0, A > 0 B = 0 single line through 0. (8) K = 0, A < 0, B = 0 single line through 0. (9) K > 0, A = 0 B = 0 empty set. (10) K = 0 A = 0, B = 0 whole plane. 196 Exercise 8.3.5 Let 8 61/2 61/2 7 Then det(t − A) = (t − 8)(t − 7) − 6 = t2 − 15t + 50 = (t − 10)(t − 5) so A has eigenvalues 5 and 10. x y A =5 x y ⇒ 3x = 61/2 y 2y = 61/2 x so the eigenvector (1, (3/2)1/2 )T gives one axis of symmetry. A x y = 10 x y ⇒ −2x = 61/2 y −3y = 61/2 x so the eigenvector (1, −(2/3)1/2 )T gives the other axis of symmetry. 197 Exercise 8.4.2 n j =1 (i) z, z = |zj |2 is always real and positive. (ii) We have n z, z = 0 ⇔ j =1 |zj |2 = 0 ⇔ |zj |2 = 0 for 1 ≤ j ≤ n ⇔ |zj | = 0 for 1 ≤ j ≤ n ⇔ z = 0. (iii) We have n ∗ (λzj )wj λz, w = j =1 n ∗ zj wj = λ z, w . =λ j =1 (iv) We have n ∗ (zj + uj )wj z + u, w = j =1 n ∗ ∗ zj wj + uj wj = z, w + u, w . = j =1 (v) z, w ∗ = n ∗ j =1 zj wj ∗ = n ∗ ∗∗ j =1 zj wj = n ∗ j =1 zj wj = w, z . 198 Exercise 8.4.3 The result is trivial if z = 0 or w = 0, so we need only consider the case when z, w = 0. Suppose first that z, w is real and positive. Then, if λ is real, (λz + w), (λz + w) = λ2 z, z + 2λ z, w + w, w = λ z, z 1/2 + z, w z, z 1/2 + w, w − z, w z, z 2 Thus, taking λ = λ0 , with λ0 = ( z, w )1/2 ( z, z )−1/2 we see that w, w − z, w z, z 2 ≥0 with equality if and only if (λ0 z + w), (λ0 z + w) = 0 and so if and only if λ0 z + w = 0 In general, we may choose θ so that eiθ z, w is real and positive. Then the result above, applied to eiθ z, yields z, w 2 w, w − e2iθ ≥0 z, z so | z, w | ≤ z w . with equality only possible if λ1 z + w = 0 for some λ1 ∈ C. By inspection, if this last condition holds, we do have equality. 199 Exercise 8.4.4 (i) z ≥ 0 since we take the positive square root. (ii) z = 0 ⇔ z, z = 0 ⇔ z = 0. (iii) λz 2 = λz, λz = λλ∗ z, z = (|λ| z )2 , so λz = |λ| z . (iv) We have z+w 2 = z + w, z + w =z 2 + z, w + w, z + w =z 2 ≤z 2 + 2ℜ z, w + w +2 z w + w = ( z + w )2 The result follows on taking square roots. 2 2 2 200 Exercise 8.4.6 (i) Observe that n j =1 n λj ej = 0 ⇒ λj ej , ek j =1 = 0 ∀k n ⇒ j =1 λj δjk = 0 ∀k ⇒ λk = 0 ∀k, so we have a set of n linearly independent vectors in a space of dimension n which is thus a basis. (ii) By (i), n z= λj ej j =1 for some λj . We observe that n λj = n λj δjk = j =1 λj ej , ek = z, ek j =1 (iii) False. If n > m, m vectors cannot span Cm . 201 Exercise 8.4.7 If k = q , we are done, since a linearly independent set forms a basis if and only if the number of elements in the basis equals the dimension of the space. If not, then k < q and there exists a so u ∈ U \ span{e1 , e2 , . . . , ek } k v =u− u, ej ej j =1 is a non-zero element of U . Setting ek+1 = v ek+1 orthonormal in U . −1 v we have e1 , e2 , . . . , After repeating the process at most q times we obtain a basis for U . 202 Exercise 8.4.8 (i) Observe that 2 k z− λj ej , k =z 2 j =1 − k λj ej , z j =1 ∗ − k λ∗ ej , z + j j =1 j =1 k k ∗ = j =1 (λj − ej , z )(λj − ej , z ) + z k = j =1 |λj |2 2 − j =1 k |λj − ej , z |2 + z 2 − j =1 | ej , z |2 k ≥z 2 − j =1 | ej , z |2 with equality if and only if λj = z, ej for all j . (ii) Since k z= z, ej ej j =1 if and only if z ∈ span{e1 , e2 , . . . , ek }, it follows from (i) that k z 2 ≥ z, ek 2 , j =1 with equality if and only if z ∈ span{e1 , e2 , . . . , ek }. | ej , z |2 203 Exercise 8.4.9 We have z+w =z 2 − z−w 2 2 = z + w, z + w − z − w, z − w + z, w − z, w ∗ +w = 4ℜ z, w . 2 −z 2 + z, w + z, w ∗ −w Thus z + iw and z+w 2 2 − z − iw − z−w 2 2 = 4ℜ z, iw = ℑ z, w + i z + iw 2 − i z − iw 2 = 4 z, w . 2 204 Exercise 8.4.10 Existence Choose an orthonormal basis ej . If α has matrix A = (aij ) with respect to this basis, let α∗ be the linear map with matrix A∗ = (bij ) where bij = a∗ . Then ji n α zj ej , j =1 n n n wj e r aij ei , zj = r =1 n = wr e r r =1 i=1 n j =1 n n zj wr aij δir j =1 i=1 r =1 n n = zj wi aij j =1 i=1 n n zj wi b∗ ji = i=1 j =1 n = n zj ej , α j =1 ∗ wj e r , r =1 so αz, w = z, α∗ w for all z, w ∈ Cn . Uniqueness Observe that, if β and γ are linear maps with αz, w = z, β w = z, γ w for all z, w ∈ Cn , then (β − γ )w = 0 for all w ∈ Cn (set z = (β − γ )w), so β = γ . The required formula A∗ = (bij ) with bij = a∗ follows from the ji results already established in the question. 205 Finally n ∗ ∗ det α = det A = i=1 σ ∈Sn n = a∗ − 1 (j ) jσ ζ (σ ) j =1 n σ ∈Sn = a∗ (i)i σ ζ (σ ) a∗ (j ) jτ ζ (σ ) j =1 τ ∈Sn ∗ n = ζ (σ ) ajτ (j ) j =1 τ ∈Sn ∗ = (det A) = (det α)∗ 206 Exercise 8.4.11 (i)⇒(ii) If (i) holds, then using the polarisation identity 4 αz, αw = αz + αw 2 = α(z + w) 2 = z+w = 4 z, w , 2 − αz − αw − α(z − w) − z−w 2 2 + i αz + iαw 2 + i α(z + iw) + i z + iw 2 2 − i αz − iαw 2 − i α(z − iw) − i z − iw 2 so (ii) holds. (ii)⇒(iii) if (ii) holds α∗ αz, w = w, α∗ αz ∗ = αw , αz ∗ = αz, αw = z, w for all w so α∗ αz = z for all z so α∗ α = ι. (iii)⇒(iv) Immediate. (iv)⇔(v)⇔(vi) Immediate. (iv)⇒(i) If (iv) holds, αz 2 = αz, αz = z, α∗ αz = z, z = z 2 . 2 2 207 Exercise 8.4.12 (i) U (Cn ) is a subset of the group GL(Cn ). We have α, β ∈ U (Cn ) ⇒ (αβ )∗ (αβ ) = (β ∗ α∗ )(αβ ) = β ∗ (α∗ α)β = β ∗ ιβ = β ∗ β = ι ⇒ αβ ∈ U (Cn ) and α ∈ U (Cn ) ⇒ (α−1 )∗ = α∗∗ = α = (α−1 )−1 ⇒ α−1 ∈ U (Cn ) whilst ι∗ ι = ιι = ι so ι ∈ U (Cn ), Thus U (Cn ) is a subgroup of GL(Cn ). (ii) 1 = det ι = det αα∗ = det α det α∗ = det α(det α)∗ = | det α|2 . The converse is false. If A= 10 , 11 then det A = 1, but A−1 = 11 01 = A∗ . 208 Exercise 8.4.13 If D is a real diagonal matrix with diagonal entries dj , D ∈ O(Rn ) ⇔ DDT = I ⇔ d2 = 1 ∀j ⇔ dj = ±1 ∀j. j If D is a complex diagonal matrix with diagonal entries dj , D ∈ U (Cn ) ⇔ DD∗ = I ⇔ dj d∗ = 1 ∀j ⇔ |dj | = ±1 ∀j. j The diagonal matrix with first diagonal entry eiθ and all other diagonal entries 1 is unitary with determinant eiθ . 209 Exercise 8.4.14 SU (Cn ) is a subset of the group SU (Cn ) containing ι. α, β ∈ SU (Cn ) ⇒ det αβ = det α det β = 1 × 1 = 1 ⇒ αβ =∈ SU (Cn ). α ∈ SU (Cn ) ⇒ det α−1 = (det α)−1 = 1 ⇒ α−1 =∈ SU (Cn ). Thus SU (Cn ) is a subgroup of U (Cn ). 210 Exercise 8.4.15 If e1 , e2 , . . . , en is an orthonormal basis for Cn and α has matrix A with respect to this basis. A = A∗ ⇔ α = α∗ ⇔ z, αw = z, αw ∀w, z ∈ Cn ⇔ z, α∗ w = z, αw ∀w, z ∈ Cn 211 Exercise 8.4.16 If A is Hermitian, det A = det A∗ = (det A)∗ If 11 , 01 then det A = 1 is real, but A is not Hermitian. A= 212 Exercise 8.4.17 (i) If u is an eigenvector with eigenvalue λ then λu 2 = λu, u = αu, u = u, αu = u, λu = λ∗ u 2 so λ = λ∗ . (ii) If u, v are eigenvectors with distinct eigenvalues λ and µ then λ u, v = λu, v = αu, v = u, αv = u, µv = µ∗ u, v = µ u, v so u, v = 0. 213 Exercise 8.4.18 Part (ii) is equivalent to part (i) by the basis change rule. We prove part (i) by induction on n. If n = 1, then, since every 1 × 1 matrix is diagonal, the result is trivial. Suppose now that the result is true for n = m and that α : Cm+1 → Cm+1 is a symmetric linear map. We know that the characteristic polynomial must have a root and that all its roots are real. Thus we can can find an eigenvalue λ1 ∈ R and a corresponding eigenvector e1 of norm 1. Consider the subspace e⊥ = {u : e1 , u = 0}. 1 We observe (and this is the key to the proof) that u ∈ e⊥ ⇒ e1 , α u = α e 1 , u = λ1 e1 , u = 0 ⇒ α u ∈ e ⊥ . 1 1 Thus we can define α|e⊥ : e⊥ → e⊥ to be the restriction of α to e⊥ . 1 1 1 1 We observe that α|e⊥ is symmetric and e⊥ has dimension m so, by the 1 1 inductive hypothesis, we can find m orthonormal eigenvectors of α|e⊥ 1 in e⊥ . Let us call them e2 , e3 , . . . ,em+1 . We observe that e1 , e2 , . . . , 1 em+1 are orthonormal eigenvectors of α and so α is diagonalisable. The induction is complete. 214 Exercise 8.4.19 We note that det(tI − A) = (t − 5)(t − 2) − 4 = t2 − 7t + 6, so the eigenvalues of A are 7 ± 5i . 2 Eigenvectors (z, w)T corresponding to the eigenvalue by 7+5i 2 are given 10z + 4iw = (7 + 5i)z that is to say −4iz + 4w = (7 + 5i)w, (3 − 5i)z + 4iw = 0 −4iz − (3 + 5i)w = 0, so an eigenvector of norm 1 is given by 2−1/2 5−1 (4i, 5i − 3)T . Eigenvectors (z, w)T corresponding to the eigenvalue by 7−5i 2 are given 10z + 4iw = (7 − 5i)z that is to say, −4iz + 4w = (7 − 5i)w, (3 + 5i)z + 4iw = 0 −4iz − (3 − 5i)w = 0, so an eigenvector of norm 1 is given by 2−1/2 5−1 (5i − 3, −4i)T . Thus U = 2−1/2 5−1 is a unitary matrix with U ∗ AU = 4i 5i − 3 5i − 3 −4i (7 + 5i)/2 0 0 (7 − 5i)/2 215 Exercise 8.4.20 We can do this using matrices, but it is nicer to use maps. If γ is unitary, set α = (γ + γ ∗ )/2, β = (γ − γ ∗ )/(2i). Then γ = α + iβ and α∗ = (γ ∗ + γ ∗∗ )/2 = (γ ∗ + γ )/2 = α β ∗ = −(γ ∗ + γ ∗∗ )/(2i) = −(γ ∗ − γ )/(2i) = β so α and β are Hermitian. To prove uniqueness, suppose that γ is unitary, α and β Hermitian and γ = α + iβ . Then so α − iβ = α∗ − iβ ∗ = (α + iβ )∗ = γ ∗ = γ 2α = (α − iβ ) + (α + iβ ) = 2γ and α = (γ + γ )/2, iβ = γ − α. ∗ We observe that αβ = (γ + γ ∗ )(γ − γ ∗ )/(4i) = (γ 2 − 2ι + (γ ∗ )2 )/(4i) and (γ − γ ∗ )(γ + γ ∗ )/(4i) = βα α2 + β 2 = 4−1 ((γ + γ ∗ )2 − (γ − γ ∗ )2 ) = 4−1 (2γγ ∗ + 2γ ∗ γ ) = ι. If we look at 1×1 matrices we get G = (eiθ ), A = (cos θ), B = (i sin θ) and recover Euler’s formula. 216 Exercise 9.1.2 We have LLT = I and det L = det 2−1/2 −2−1/2 2−1/2 2−1/2 so L ∈ SO(R3 ). =1 However, if we set x = (0, 1, 0)T , we have x′ = (0, 2−1/2 , 2−1/2 ), so 3 x′24 −1 =4 −1/2 =2 = l2j x4 j l2 x4 2 j =1 217 Exercise 9.1.4 (i) We have d′ d ui = lij uj = lij uj ˙ dt dt ˙ so u is a Cartesian tensor of order 1. u′i = ˙ ˙ (ii) x is a Cartesian tensor of order 1, so x is, so ¨ is, so F = m¨ is. x x 218 Exercise 9.2.1 (i) Observe that ′ aij = u′i vj = lir ur ljs vs = lir ljs ars . (ii) Observe that ∂u′j ∂u′j ∂xr ∂ljk uk ∂uk a′ij = = = lir = lir ljk = lir ljk ark . ′ ′ ∂xi ∂xr ∂xi ∂xr ∂xr (iii) Observe that a′ij = ∂ 2 φ ∂xr ∂xs ∂ 2φ = = lir ljs ars . ∂x′i ∂x′j ∂xr ∂xs ∂x′i ∂x′j 219 Exercise 9.3.1 (i) If ui , vj wk are Cartesian tensors of order 1 then ′′ (ui vj wk )′ = u′i vj wk = lir ljs lkp ur vs wp so ui vj wk is a Cartesian tensor of order 3. (ii) If uij is a Cartesian tensor of order 2 then ∂u′ij ∂lir ljs u′rs ∂xp ∂urs = = lir ljs lkp ′ ′ ∂xk ∂xp ∂xk ∂xp so ∂uij ∂xk is a Cartesian tensor of order 3. (iii) If φ is a smooth tensor of order 0 ∂2φ ∂ 3φ ∂xr ∂xs ∂xp ∂3φ = = lir ljs lkp ∂x′i ∂x′j ∂x′k ∂xr ∂xs ∂xp ∂x′i ∂x′j ∂x′k ∂xr ∂xs ∂xp so ∂ 2φ ∂xi ∂xj ∂xk is a Cartesian tensor of order 3. 220 Exercise 9.3.5 (i) Observe that, in our standard notation lik a′i bk = a′i b′i = (ai bi )′ = ai bi = ak bk . and so (lik a′i − ak )bk = 0. Since we can assign bk any values we please in a particular coordinate system, we must have lik a′i − ak = 0, so δir a′i − lrk ak = lrk (lik a′i − ak ) = 0 and we have shown that a′r = lrk ak and ai is a Cartesian tensor of order 1. (ii) By (i), aij bi is a Cartesian tensor of rank 1 whenever bi is a Cartesian tensor of order 1 and so, by Theorem 9.3.4, aij is a Cartesian tensor of order 2. (iii) Apply (ii) with uij = bi cj . (iv) Observe that, in our standard notation lkr lms a′ijkm brs = a′ijkm b′km = (aijkm bkm )′ = lip ljq (apqkm bkm ) = lip ljq (apqrs brs ) and so (lkr lms a′ijkm − lip ljq apqrs )brs = 0. Since we can assign brs any values we please in a particular coordinate system, we must have so whence a′ijef sian tensor. lkr lms a′ijkm − lip ljq apqrs = 0. ler lf s (lkr lms a′ijkm − lip ljq )apqrs = 0 = ler lf s lip ljqapqrs and we have shown that aijkl is a Carte- 221 Exercise 9.3.6 (i) We have 1 cijkm ekm = 2 (˜ijkm + cijmk )ekm = 1 (˜ijkm ekm + cijmk ekm ) c ˜ c ˜ 2 1 c ˜ = 2 (˜ijkm ekm + cijmk emk ) = pij and 1 cijmk = 1 (˜ijmk + cijkm ) = 2 (˜ijkm + cijmk ) = cijkm . c ˜ c ˜ 2 1 1 (ii) Set emk = 2 (bkm + bmk ) and fmk = 2 (bkm − bmk ). (iii) We have cijmk fmk = cijkm fkm = −cijmk fmk so cijmk fmk = 0 and cijkm ekm = cijmk bmk . Thus cijmk bmk is a Cartesian tensor of order 2 whenever bmk is, so, by the quotient rule given in Exercise 9.3.5 (iv), cijmk is a Cartesian tensor of order 4. 222 Exercise 9.3.8 If we work with a fixed set of coordinates, Cartesian tensors of rank n may be identified with the space of functions f : R3n → R with pointwise multiplication and multiplication by scalars. They therefore form a vector space of dimension 3n . 223 Exercise 9.4.3 Using the Levi-Civita identity, we have ǫijk ǫklm ǫmni = ǫkij ǫklm ǫmni = (δil δjm −δim δjl )ǫmni = ǫjnl −ǫmnm = ǫjnl = ǫljn . 224 Exercise 9.4.4 Observe that ζ (σ )2 = ǫijk...p ǫijk...p = σ ∈Sn 1 = card Sn = n! σ ∈Sn (See the definition of the determinant on page 87 if necessary.) 225 Exercise 9.4.5 (i) We have λ(ǫijk aj bk ) = ǫijk (λaj )bk = ǫijk aj (λbk ), so λ(a × b) = (λa) × b = a × (λb). (ii) We have ǫijk aj (bk + ck ) = ǫijk aj bk + ǫijk aj ck , so a × (b + c) = a × b + a × c. (iii) We have ǫijk aj bk = −ǫikj aj bk = −ǫikj bk aj , so a × b = −b × a. (iv) Putting b = a in (iii), we get so a × a = 0. a × a = −a × a 226 Exercise 9.4.6 There is no identity. For suppose e was an identity. Then e = e × e = 0, but 0×a=0 for all a so (since R has more than one element) 0 is not an identity. 3 227 Exercise 9.4.7 (i) We have λ(aj bj ) = (λaj )bj = aj (λbj ), so λ(a · b) = (λa) · b = a · (λb). (ii) We have ak (bk + ck ) = ak bk + ak ck , so a · (b + c) = a · b + a · c. (iii) We have aj b j = b j aj , so a · b = b · a. 228 Exercise 9.4.9 (i) We observe that a × (b × c) = (a1 , a2 , a3 ) × (b2 c3 − b3 c2 , b3 c1 − b1 c3 , b1 c2 − b2 c1 ) = a2 (b1 c2 − b2 c1 ) − a3 (b3 c1 − b1 c3 ), a3 (b2 c3 − b3 c2 ) − a1 (b1 c2 − b2 c1 ), a1 (b3 c1 − b1 c3 ) − a2 (b2 c3 − b3 c2 ) = (a2 c2 + a3 c3 )b1 , (a3 c3 + a1 c1 )b2 , (a1 c1 + a2 c2 )b3 − (a2 b2 + a3 b3 )c1 , (a3 b3 + a1 b1 )c2 , (a1 b1 + a2 b2 )c3 = (a2 c2 + a3 c3 + a1 c1 )b1 , (a3 c3 + a1 c1 + a2 c2 )b2 , (a1 c1 + a2 c2 + a3 c3 )b3 − (a2 b2 + a3 b3 + a1 b1 )c1 , (a3 b3 + a1 b1 + a2 b2 )c2 , (a1 b1 + a2 b2 + a3 b3 )c3 = (a · c)b − (a · b)c. (ii) ‘Search me, guv’ner!’ There is a geometric proof in the American Mathematical Monthly (Vol 72, Feb 1965) by K. Bishop, but I doubt if the reader will call it simple. 229 Exercise 9.4.10 We have (a × b) × c = −c × (a × b) = − (c · b)a − (c · a)b = (a · c)b − (b · c)a. as required. If x = (1, 0, 0), y = z = (0, 1, 0) then (x × y) × z = (0, 0, 1) × (0, 1, 0) = (−1, 0, 0) = (0, 0, 0) = (0, 0, 1) × (0, 0, 0) = x × (y × z). 230 Exercise 9.4.11 We have a × (b × c) + b × (c × a) + c × (a × b) = (a · c)b − (a · b)c + (b · a)c − (b · c)a + (c · b)a − (c · a)b = 0. 231 Exercise 9.4.12 We have (a × b) · (a × b) + (a · b)2 = ǫijk aj bk ǫirs ar bs + ai bi aj bj = (δjr δks − δjs δkr )aj bk ar bs + ai bi aj bj = a j aj b k b k − aj b j ak b k + ai b i aj b j = a j aj b k b k = a 2 b 2. If we use the geometric formulae for a × b and a · b, our formula becomes a2 b2 cos2 θ + a2 b2 sin2 θ = a2 b2 where θ is the angle between a and b, a = a and b = b . We thus recover the formula cos2 θ + sin2 θ = 1. 232 Exercise 9.4.13 (i) We have so (a × b) · a = ǫijk aj bk ai = −ǫjik aj bk ai = −(a × b) · a (a × b) · a = 0 (ii) a × b ⊥ a so (a × b) · a = 0 (iii) The volume of a parallelepiped with one vertex at 0 and adjacent vertices at a, b and a is 0 so (a × b) · a = [a, b, a] = 0. 233 Exercise 9.4.14 We have (a × b) × (c × d) = ((a × b) · d)c − ((a × b) · c)d We also have = [a, b, d]c − [a, b, c]d. (a × b) × (c × d) = −(c × d) × (a × b) = (c × d) · ab − c × d) · ba = [c, d, a]b − [c, d, b]a = −[c, a, d]b − [b, c, d]a. Thus a, b, d]c − [a, b, c]d = [c, a, d]b − [b, c, d]a and the required result follows. Our formulae also give (a × b) × (a × c) = ((a × b) · c)a − ((a × b) · a)c since (a × b) ⊥ a = [a, b, c]a − 0 = [a, b, c]a 234 Exercise 9.4.15 Since b × c ⊥ b, c we have so x · (b × c = (λa + µb + ν c) · (b × c) = λa · (b × c) λ= [x, b, c] . [a, b, c] Similarly µ= [a, b, x] [a, x, b] and ν = . [a, b, c] [a, b, c] 235 Exercise 9.4.16 (i) We work with row vectors 2 a [a, b, c]2 = det b c a b det aT = det c T aa abT = det baT bbT caT cbT a·a a·b b · a b · b = det c·a c·b bT cT ) (ii) In the case given we first observe that r2 = a + b + c 2 =a 2 +b 2 +c so 2 acT bcT ccT a·c b · c c·c + 2(a · b + b · c + a · b a · b + b · c + a · b = −r2 a·a [a, b, c]2 = det b · a c·a 2 r = det b · a c·a a·b a·c b · b b · c c·b c·c a·b a·c r2 b · c c · b r2 = r2 (r4 − (b · c)2 ) − a · b(r2 a · b − (b · c)(a · c)) + a · c((b · a)(b · c) − r2 b · a) = 2r6 + 2r4 (a · b + b · c + a · b) + r2 (a · b + b · c + a · b)2 − r2 ((a · b)2 + (b · c)2 + (c · a)2 ) + 2(a · b)(b · c)(c · a) = 2(r2 + a · b)(r2 + b · c)(r2 + c · a). 236 Exercise 9.4.17 Observe, using the summation convention, that (a × b) · (c × d) = ǫijk aj bk ǫirs cr ds = (δjr δks − δkr δjs )aj bk cr ds = aj cj bk dk − aj dj bk ck = (a · c)(b · d) − (a · d)(c · d) (Or we could use the formula for the triple vector product.) Thus (a × b) · (c × d) + (a × c) · (d × b) + (a × d) · (b × c) = (a · c)(b · d) − (a · d)(c · d) + (a · d)(b · c) − (a · b)(c · d) + (a · b)(c · d) − (a · c)(b · d) = 0. 237 Exercise 9.4.19 (i) Using the summation convention, we have d φa dt = i d ˙ ˙ ˙ (φai ) = φai + φai = φa + φa ˙ dt (ii) Using the summation convention, we have d d ˙ ˙ ˙ a · b = ai b i = ai b i + ai b i = a · b + a · b . ˙ dt dt (iii) By Lemma 9.4.18, d ˙ ˙˙ ¨ ¨ a × a = a × a + a × a = a × a. dt i . 238 Exercise 9.4.21 (i) Using the summation convention, we have d ∇ × (φu) i = ǫijk φuk dxj duk dφ uk + ǫijk φ = ∇φ) × u + φ∇ × u i . = ǫijk dxj dxj (ii) Using the summation convention, we have duk d2 uk d2 uk d ǫijk = ǫijk = ǫijk = −∇ · (∇ × u) ∇ · (∇ × u) = dxi dxj dxi dxj dxj dxi so ∇ · (∇ × u) = 0. (ii) Using the summation convention, we have d ∇ × (u × v) i = ǫijk ǫkrs ur vs dxj d ur v s = ǫkij ǫkrs dxj d ur v s = (δir δjs − δis δjr ) dxj d d ui v j − uj v i = dxj dxj dvj duj dvi dui + ui − vi − uj = vj dxj dxj dxj dxj = (∇ · u)v + u · ∇v − (∇ · v)u − v · ∇u i . 239 Exercise 9.4.22 (i) Observe that the cancellation law continues to hold, so it is sufficient to show that the analogue of the scalar vector product (think of the 4 dimensional volume) is a scalar. Let ai = ǫijkl xj yk zl . Then if bi is an vector (ie order 1 Cartesian tensor), ′ (ai bi )′ = ǫijkl x′j yk zl′ b′i = ǫijkl lju lkv llv lip xu yv zw bp = (det L)ǫpuvw xu yv zw bp = ǫpuvw xu yv zw bp = ai bi since L ∈ SO(R4 ). Thus, by the cancellation law, x ∧ y ∧ z is a vector (ii) We have (λx + µu) ∧ y ∧ z i = ǫijkl (λxj + µuj )yk zl = λǫijkl xj yk zl + µǫijkl uj yk zl = (λx ∧ y ∧ z + µu ∧ y ∧ z)i . so (y ∧ x ∧ z)i = ǫijkl yj xk zl = −ǫikjl xk yj zl = (x ∧ y ∧ z)i The proof that follows the same lines. x ∧ y ∧ z = −y ∧ x ∧ z. x ∧ y ∧ z = −z ∧ y ∧ x (iii) Using the summation convention with range 1, 2, 3, 4, 5, we define x∧y∧z∧w =a with ai = ǫijklu xj yk zl wu . 240 Exercise 10.1.4 Since the material is isotropic so is aijk thus aijk = κǫijk and σij = aijk Bk = κǫijk Bk = −κǫjik Bk = −σji = −σij . Thus σij = 0. 241 Exercise 10.1.2 It is sufficient to show that δij δkl is isotropic, since it then follows that δik δjl and δil δjk and all linear combinations are. ′′ But (δij δkl )′ = δij δkl = δij δkl so we are done. If αδij δkl + βδik δjl + γδil δjk = 0, then 0 = αδij δkk + βδik δjk + γδik δjk = (3α + β + γ )δij . Thus 3α + β + γ = 0 and similarly α + 3β + γ = 0, α + β + 3γ = 0. Thus 2α = 2β = 2γ = −(α + β + γ ) and α = β = γ = 0. (Alternatively, just examine particular entries in the tensor αδij δkl + βδik δjl + γδil δjk .) For the last part, observe that, since ǫijk and ǫist are isotropic, so is ǫijk ǫist . Thus ǫijk ǫist = αδjk δst + βδjs δkt + γδjt δks. We observe that αδjk δst +βδjs δkt +γδjt δks = ǫijk ǫist = −ǫikj ǫist = −αδjk δst +γδjs δkt +βδjt δks so, by the linear independence proved earlier, β = −γ . Next we observe that 0 = ǫijj ǫist = (3α + β + γ )δst so 3α + β + γ = 0 and α = 0 Finally 6 = ǫijk ǫijk = 3α + 9β + 3γ so α = 0, β = 1 and γ = −1. We have shown that ǫijk ǫist = δjs δkt − δks δjt . 242 Exercise 10.1.3 a′ij = lir ljs ars = lir ljs asr = a′ji . 243 Exercise 10.1.6 (i) We have a′ij = lir ljs ars = −lir ljs asr = −a′ji . (ii) We have 1 bij = 2 (bij + bji ) + 1 (bij − bji ) 2 and 1 (b 2 ji 1 (b 2 ji 1 + bij ) = 2 (bij + bji ) − bij ) = − 1 (bij − bji ). 2 (iii) Observe that the zero order two tensor is both symmetric and antisymmetric. Further and aij = aji , bij = bji ⇒ λaij + µbij = λaji + µbji aij = −aji , bij = −bji ⇒ λaij + µbij = −(λaji + µbji ) Thus the symmetric tensors form a subspace and the antisymmetric tensors form a subspace. We do not use the summation convention for α and β . Let e(α, β ) be the second order Cartesian tensor with eij (α, β ) = δαi δβj + δαi δβj in a particular coordinate system S [1 ≤ α < β ≤ 3] and eij (α, α) = δαi δαj for 1 ≤ α ≤ 3 Let f (α, β ) be the second orderCartesian tensor with fij (α, β ) = δαi δβj − δαi δβj in the same coordinate system S [1 ≤ α < β ≤ 3]. Then in that coordinate system e(α, β ) is symmetric, f (α, β ) is antisymmetric. If a is a second order symmetric tensor then in the coordinate system S , a= aαβ e(α, β ) 1≤α≤β ≤3 and if b is an antisymmetric tensor then in the coordinate system S , b= bαβ f (α, β ). 1≤α<β ≤3 Since the result is true in one system it is true in all. 244 Working in S we see that 1≤α≤β ≤3 aαβ e(α, β ) = 0 ⇒ aαβ = 0 [1 ≤ α ≤ β ≤ 3] and 1≤α<β ≤3 bαβ e(α, β ) = 0 ⇒ bαβ = 0 [1 ≤ α < β ≤ 3]. Thus the e(α, β ) give a basis for the subspace of symmetric tensors which thus has dimension 6 the f (α, β ) give a basis for the subspace of antisymmetric tensors which thus has dimension 3. 245 Exercise 10.1.8 Since ǫijk ωj xk = (ω × x)i and ω × x ⊥ x, the only possible eigenvalue is 0 and this corresponds to eigenvectors tω with t = 0. 246 Exercise 10.2.1 Observe first that m α rα = α α mα xα − M xG = 0. We have H= α = α = α ˙ mα xα × xα ˙ mα (rα + xG ) × (˙ α + xG ) r ˙ mα rα × rα + xG × + m α rα α = α = α ˙ m α rα + α α × xG ˙ mα rα × rα + xG × d dt ˙ mα rα × rα + xG × d 0+ dt ˙ = M xG × xG + α ˙ m α rα × rα m α rα + α α ˙ m α r α × r α + 0 × xG m α rα α ˙ m α rα × rα The total angular momentum about the origin is the angular momentum of the system around the centre of mass plus the angular momentum of a point mass equal to the mass of the system at the centre of gravity around the origin. 247 Exercise 10.2.2 We have d ˙ H= dt = α = α = α = α = −k α ˙ mα xα × xα ˙ ˙ ¨ mα (xα × xα + xα × xα ) ¨ mα xα × xα ¨ xα × mα xα xα × Fα α = −k H ˙ mα xα × xα Thus (you can work with the summation convention, if you prefer) d kt e H(t) = 0 dt so H(t)ekt is a constant vector H0 and we are done. 248 Exercise 10.2.3 Since the first lump has centre of mass at 0 we have xi ρ(x) dV (x) = 0 By a simple change of variable Jij = = (xk + ak )(xk + ak )δij − (xi + ai )(xj + aj ) ρ(x) dV (x) (xk xk δij − xi xj )ρ(x) dV (x) + 2ak δij + (ak ak δij − ai aj ) + aj ρ(x) dV (x) + ai xi ρ(x) dV (x) = Iij + M (ak ak δij − ai aj ). xk ρ(x) dV (x) xj ρ(x) dV (x) 249 Exercise 10.2.4 In the coordinate system chosen, taking x1 = x, x2 = y , x3 = z , A = I11 = (xk xk δ11 − x1 x1 )ρ(x) dV (x) = (x2 x2 + x3 x3 )ρ(x) dV (x) = (y 2 + z 2 )ρ(x, y, z ) dx dy dz ≥ 0 B= (z 2 + x2 )ρ(x, y, z ) dx dy dz ≥ 0, C= (x2 + y 2 )ρ(x, y, z ) dx dy dz ≥ 0. We also have 250 Exercise 10.3.2 (i) We have lir ljs δrs = lir ljr = δij T since LL = I . (ii) If aijk is an isotropic extended Cartesian tensor, then aijk is an isotropic Cartesian tensor so aijk = Aǫijk so aijk = 0. (iii) We have (ǫrst ar bs ct )′ = lri lsj ltk ǫrst ai bj ck = (det L)ǫijk ai bj ck = −ǫijk ai bj ck . (iv) We have (ar br )′ = a′r b′r = lri lrj ai bj = δij ai bj . (v) If a × b was an extended Cartesian tensor then by (iv) so would be (a × b) · c. Now choose c non-zero and orthogonal to a and b. 251 Exercise 10.4.1 (i) We have i i i (λui + µv i )¯ = (λui + µv i ) = (λlj uj + µlj v j ) = lj (λuj + µv j ). ¯ ¯ (ii) We have i i (¯′ )i = (lj ui )′ = lj uj . u ˙ 252 Exercise 10.4.3 i mi = lj ⇔ L = (LT )−1 ⇔ LT = L−1 ⇔ L ∈ O(R3 ). j 253 Exercise 10.4.6 (i) We have j i i v ¯v lu mv δi = lu mv = δu = δu . j i Choose an L such that L2 = I , for example 10 0 4 L = 0 5 − 3 . 5 3 05 4 5 Then uv uv ¯ li lj δuv = li lu = δij = δij . (ii) The rule is uv akn = li lj mk mn apq . ¯ij p q uv (iii) We have uv uv kn uv u v uv li lj mk mn apq = li lj mk mn δi δj = li lj mi mj = li mi lj mj = δp δq = auv , ¯pq p q uv pq pq p q so akn is a 4th rank tensor. ij Next observe that n in ain = δi δj = 3δj ij a second rank tensor. However kn akn = δi δi = δkn ii is not a second rank tensor . The same argument shows that ann is ij not. (iv) The rule is uvp an = li lj lk mn aq . ¯ijk q uvp We observe that uvp uv p uv an = li lj ln mn aq = li lj δq aq = li lj ap ¯ijn q uvp uvp uvp so an is a second rank tensor. ijn 254 Exercise 11.1.2 Observe that n (α + β )(ej ) = α(ej ) + β (ej ) = n aij fi + i=1 and n λα(ej ) = λ bij fi = i=1 n aij fi = i=1 n λaij fi . i=1 (aij + bij )fi i=1 255 Exercise 11.1.3 If p aij gi α(fj ) = i=1 n β (er ) = bsr fs , s=1 then n αβ (er ) = s=1 n ais bsr ais gi = bsr s=1 p p n bsr αfs = i=1 i=1 s=1 so αβ has matrix AB with respect to the given bases. gi , 256 Exercise 11.1.4 (i) Observe that, by definition, (α + β )γ u = α(γ u) + β (γ u) = (αγ )u + (βγ )u = (αβ + αγ )u for all u ∈ U so, by definition (α + β )γ = αγ + βγ. (ii) and (iii) follow at once from Exercises 11.1.2 and 11.1.3. (iv) Let U , V and W be (not necessarily finite dimensional) vector spaces over F and let α, β ∈ L(U, V ), γ ∈ L(V, W ). Then γ (α + β ) = γα + γβ. If A and B are n × m matrices over F and C is an p × n matrix over F, we have C (A + B ) = CA + CB. 257 Exercise 11.1.5 Since the fi are a basis for V , we can find pij ∈ F such that n fj′ pij fi [1 ≤ j ≤ n] = i=1 and, similarly, we can find mij and qrs such that n mij fi [1 ≤ j ≤ n] fj = i=1 and m e′s = r =1 qrs er [1 ≤ s ≤ m]. We have n n mij fi′ = fj = i=1 n mij i=1 pki fk k=1 so by uniqueness n δkj = pki mij i=1 that is to say P M = I so P is invertible with M = P −1 . Similarly m αe′s = α qrs er r =1 m = qrs αer r =1 m = n ′ apr fp qrs p=1 n r =1 m = qrs r =1 m mip fi′ apr p=1 n n i=1 m mip apr qrs i=1 p=1 r =1 so B = M AQ = P −1 AQ. ′ fp , 258 Exercise 11.1.9 (i) Since I ∈ G, A ∈ X ⇒ A, A ∈ X, A = AI ⇒ A ∼1 A. If A ∼1 B then A, B ∈ X and we can find a P ∈ G such that B = P A. But P −1 ∈ G and A = P −1 B so B ∼1 A. If A ∼1 B , B ∼1 C , then A, B, C ∈ X and we can find a P, Q ∈ G such that B = P A, C = QB . Since QP ∈ G and C = (QP )A we have A ∼1 C . Thus ∼1 is an equivalence relation on X . (ii) Since I ∈ G, A ∈ X ⇒ A, A ∈ X, A = I −1 AI ⇒ A ∼2 A. If A ∼2 B , then A, B ∈ X and we can find P, Q ∈ G such that B = P −1 AQ. But P −1 , Q ∈ G and A = (P −1 )−1 BQ−1 , so B ∼2 A. If A ∼2 B , B ∼2 C , then A, B, C ∈ X and we can find P, Q, R, S ∈ G such that B = P −1 AQ, C = R−1 BS . Since P R, SQ ∈ G and C = (P R)−1 A(SQ), we have A ∼2 C . Thus ∼2 is an equivalence relation on X . (iii) If A ∼2 B , then A, B ∈ X and we can find P, Q ∈ GL(Fn ) with B = P −1 AQ and so rank B = rank A. If rank B = rank A = k , then we can find P, Q, R, S ∈ GL(Fn ) such that D = P −1 AQ and D = R−1 BS where D is the n × n diagonal matrix with first k diagonal entries 1 and remaining entries 0 so A ∼2 D, D ∼2 B and A ∼2 B . Thus there are n + 1 equivalence classes corresponding to the matrices of rank k [0 ≤ k ≤ n]. (iv) Since I ∈ G. A ∈ X ⇒ A, A ∈ X, A = I −1 AI ⇒ A ∼3 A. If A ∼3 B then A, B ∈ X and we can find P ∈ G such that B = P AP . But P −1 ∈ G and A = (P −1 )−1 BP −1 so B ∼3 A. −1 If A ∼3 B , B ∼3 C , then A, B, C ∈ X and we can find P, Q ∈ G such that B = P −1 AP , C = Q−1 BQ. Since P Q ∈ G and C = (P Q)−1 A(P Q) we have A ∼3 C . Thus ∼3 is an equivalence relation on X. (v) Since A and P −1 AP have the same characteristic polynomial A ∼3 B ⇒ χA = χB Since whenever A is symmetric there exists an orthogonal matrix P such that P −1 AP = P T AP = D where D is a diagonal matrix with 259 the same characteristic polynomial χA = χB ⇒ A ∼3 B. Thus A ∼3 B ⇔ χA = χB . If Dλ is the diagonal matrix with first diagonal entry λ and all others 0 then Dλ ∼3 Dµ ⇔ χDλ = χDµ ⇔ λ = µ. Thus there are infinitely many equivalence classes for ∼3 . (vi) Since I ∈ G A ∈ X ⇒ A, A ∈ X, A = I T AI ⇒ A ∼3 A. If A ∼4 B , then A, B ∈ X and we can find P ∈ G such that B = P T AP . But P −1 ∈ G and A = (P −1 )T BP −1 , so B ∼4 A. If A ∼4 B , B ∼4 C , then A, B, C ∈ X and we can find P, Q ∈ G such that B = P T AP , C = QT BQ. Since P Q ∈ G and C = (P Q)T A(P Q) we have A ∼4 C . Thus ∼4 is an equivalence relation on X. (vii) If A is a real symmetric n × n matrix we can find P ∈ O(Rn ) such that D = P T AP with D = (dij ) a diagonal matrix with dii > 0 for 1 ≤ i < u, dii < 0 for u ≤ i < v and dii = 0 otherwise. Let E = (eij ) a −1/2 diagonal matrix with eii = dii for 1 ≤ i < u, eii = (−dii )−1/2 for u ≤ i < v and eii = 1 otherwise. Then P E ∈ GL(Rn ) and (P E )T A(P E ) is a diagonal matrix with entries 1, −1 and 0. Thus there can only be finitely many equivalence classes. [The reason that this does not complete the discussion is that we have not shown (as we will in Section 15.2) that different (u, v ) correspond to different equivalence classes.] 260 Exercise 11.1.15 Let U and V be finite dimensional vector spaces. and let α : U → V be linear. By Theorem 11.1.13, dim im α = dim(U/ ker α) and, by Lemma 11.1.14, dim U = dim ker α + dim(U/ ker α), so dim U = dim ker α + dim(im α). 261 Exercise 11.1.16 Observe that dim Hj = dim Bj − dim Zj = rank(αj +1 ) − (dim Cj − rank(αj )) and so dim Hj + dim Cj = rank(αj +1 ) + rank(αj ) for 1 ≤ j ≤ n Now rank(αn+1 ) = rank(α0 ) = 0 since Cn+1 = C0 = {0} so n j =1 (−1)j (dim Hj + dim Cj ) = 0. 262 Exercise 11.2.1 The sum of two polynomials is polynomial and the product of a polynomial with a real number is a polynomial, so we have a subspace of the vector space of functions f : R → R with pointwise addition and scalar multiplication. (i) Let λ, µ ∈ R, P, Q ∈ P . We have D(λP + µQ) = λDP + µDQ, by the standard rules for differentiation. We have M (λP +µQ)(t) = t λP (t)+µQ(t) = λtP (t)+µtQ(t) = (λM P +µM Q)(t), so M (λP + µQ) = λM P + µM Q. We have Eh (λP + µQ)(t) = (λP (t + h) + µQ(t + h)) = λEh P (t) + µEh Q(t) = (λEh P + µEh Q)(t), so M (λP + µQ) = λM P + µM Q. (ii) We have d tP (t) − tP ′ (t) = P (t), dt so (DM − M D)P = P for all P ∈ P and DM − M D = ι the identity map. (DM − M D))P (t) = (iii) Well defined since for each P we only consider a finite sum and if αj P = 0 for all j ≥ N N M αj P = j =0 αj P j =0 for all M ≥ N . If λ, µ ∈ R, P, Q ∈ P . we can find an N such that αj P = αj Q = 0 for all j ≥ N , so ∞ N N αj (λP + µQ) = j =0 αj (λP + µQ) = j =0 j =0 N =λ ∞ N αj P + µ j =0 (λαj P + µαj Q) αj Q = λ j =0 ∞ αj P + µ j =0 αj Q. j =0 263 Thus ∞ j =0 αj ∈ L(P , P ). (iv) If P is a polynomial of degree N , then Dj P = 0 for j ≥ N + 1. If en = tn , then M j e0 = ej = 0, so M does not have the required property. j Since Eh e0 = ej = 0, Eh does not have the required property. (ι − Eh )ek is a polynomial of degree k − 1 if k ≥ 1. and is zero if k = 0. Thus, if P is a polynomial of degree n, (ι − Eh )ek is a polynomial of degree at most n − 1 if n ≥ 1. and is zero if n = 0. Thus if P is a polynomial of degree N (ι − Eh )N +1 P = 0 and we have desired property. (v) Part (iii) tells us that exp(α) and log(ι − α) are well defined members of L(P , P ). Observe that ∞ exp α − ι = α j =1 αj −1 /(j − 1) = αβ where β is an endomorphism which commutes with α. Thus (exp α − ι)N = αN β N = 0 if αN = 0. It follows, by part (iii), that log(exp α) is well defined. By the standard rules on Taylor expansions, we have ∞ cj tj log(exp t) = j =0 for |t| small and (using those standard rules to compute cj ) ∞ cj α j . log(exp α) = j =0 However we know from calculus that log(exp t) = t for |t| small and so (by the uniqueness of Taylor expansions) c1 = 1 and cj = 0 otherwise. Thus log(exp α) = α. (vi) We know that polynomials are their own Taylor series so ∞ (exp hD)P = j =0 hj (j ) P (t) = P (t + h) = (Eh P )(t) j! and thus exp hD = Eh . 264 We know, from part (iv), that, for each P ∈ P , we can find an N (P ) such that △j (P ) = 0 for all j > N (P ). Thus (v) tells us that h log Eh = hD, that is to say ∞ (−1)j +1 j △h . j hD = j =1 (vii If P has degree N , ∞ (ι−λD) ∞ j j =0 r λ D P = (ι−λD) j =N +1 Thus λj Dr P = (ι−λN +2 DN +2 )P = P = ιP ∞ (ι − λD) λj D j = ι j =0 and similarly ∞ λj D j (ι − λD) = ι j =0 It follows that ∞ ∞ j (ι − λD)P = Q ⇒ P = λD j j =0 We also have (ι − λD)P = ∞ λj D j (ι − λD) P = P. j =0 Thus the unique polynomial solution of (ι − λD)P = Q is ∞ ∞ j Q= j λ j P (j ) . λDP= j =0 j =0 In particular, the unique polynomial solution of f ′ (x) − f (x) = x2 ie of is (ι − D)f = −e2 ∞ j =0 dj (−x2 ) = −x2 − 2x − 2. dxj λj D j j =0 Q. 265 There is only one polynomial solution. (The general, not necessarily polynomial, solution is Aex − x2 − 2x − 2 with A chosen freely.) 266 Exercise 11.2.3 If rank αn = 0, then, since rank αm = 0 for some m, rank αn − rank αn+1 ≥ 1 Thus if n ≥ r ≥ 0 so rank αr − rank αr+1 ≥ rank αn − rank αn+1 ≥ 1 n−1 n rank α ≤ n − r =0 rank αr − rank αr+1 ≤ 0 which contradicts our original hypothesis. By reductio ad absurdum, rank αn = 0. Now suppose that α has rank r and αm = 0, We have so n − r = n − rank α ≥ rank αj − rank αj +1 m m(n − r) ≥ j =0 rank αj − rank αj +1 = n − rank αm = n so mn − n ≥ mr and r ≤ n(1 − m−1 ). 267 Exercise 11.2.4 Observe that if Ak = aij (k ) is the k × k matrix with ai,i+1 (k ) = 1 for 1 ≤ i ≤ k − 1 aij (k ) = 0 otherwise, then the associated linear map αk : Fk → Fk has rank αu (k ) = max(k, u − 1). Consider the n × n matrix A which has sj − sj +1 − sj +1 − sj +2 copies of Aj and one sm × sm identity matrix along the diagonal and all other entries zero. If α has matrix A with respect to some basis, then α will have the desired property. 268 Exercise 11.2.5 If we write rk = rank αk , we know from general theorems that r0 − r1 ≥ r1 − r2 ≥ r2 − r3 ≥ . . . ≥ rn−1 − rn ≥ 0. We have r0 − r1 = 2, so 2 ≥ rj − rj +1 ≥ 0. Since n−1 2n = r0 − rn = j =0 rj − rj +1 , it follows that 2 = rj − rj +1 for 0 ≤ j ≤ n − 1 and rank αj = 2n − 2j for 0 ≤ j ≤ n. 269 Exercise 11.3.1 (i) We have δa (λ + µg ) = (λf + µg )(a) = λf (a) + µg (a) = λδa f + µδa g. (ii) We have ′ ′ ′ δa (λ + µg ) = −(λf + µg )′ (a) = −λf ′ (a) − µg ′ (a) = λδa f + µδa g. (iii) We have 1 J (λf + µg ) = (λf (x) + µg (x)) dx 0 1 1 g (x) dx f (x) dx + µ =λ 0 = λJf + µJg. 0 270 Exercise 11.3.6 Φ(α) = 0 ⇒ α′ = 0 by definition. α′ = 0 ⇒ α′ (v′ ) = 0 ∀v′ ∈ V ′ by definition. α′ (v′ ) = 0 ∀v′ ∈ V ′ ⇒ α′ (v′ )u = 0 ∀v′ ∈ V ′ ∀u ∈ U by definition. α′ (v′ )u = 0 ∀v′ ∈ V ′ ∀u ∈ U ⇒ v′ αu = 0 ∀v′ ∈ V ′ ∀u ∈ U since, by definition, α′ (v′ )u = v′ α(u). v′ αu = 0 ∀v′ ∈ V ′ ∀u ∈ U ⇒ αu = 0 ∀u ∈ U since V ′ separates V . αu = 0 ∀u ∈ U ⇒ α = 0 by definition. 271 Exercise 11.3.7 (i) We have (βα)′ w′ u = w′ (βα)u = w′ β (αu) = (β ′ w′ )(αu) = α′ (β ′ w′ ) u = (α′ β ′ )w′ u for all u ∈ U , so for all w′ ∈ W ′ , so (ii) We have for all v ∈ U ,so for all u′ ∈ U ′ , so (βα)′ w′ = (α′ β ′ )w′ (βα)′ = α′ β ′ . (ι′U u′ )v = u′ ιu v = u′ v ι′U u′ = u′ ι′U = ιU ′ . (iii) If α is invertible, then αα−1 = ι so, by the previous parts, (α−1 )′ α′ = ι so α′ is invertible and (α′ )−1 = (α−1 )′ . 272 Exercise 11.3.8 We have α′′ (Θu)v′ = Θu(α′ v′ ) = (α′ v′ )u = v ′ αu = Θ(αu) v′ for all v′ ∈ V . Since V ′′ is separated by V ′ , α′′ (Θu) = Θ(αu) for all u ∈ U . 273 Exercise 11.3.9 (i) Certainly the zero sequence 0 ∈ c00 . If a, b ∈ c00 we can find n and m such that ar = 0 for r ≥ n and br = 0 for r ≥ m. Thus if λ, µ ∈ R we have λar + µbr = 0 for all r ≥ max{n, m} so λa + µb ∈ c00 . Thus c00 is a subspace of s. (ii) If x ∈ c00 , then we can find an n such that xr = 0 for r ≥ n. We have n−1 x= xr er r =1 and n−1 Tx = n−1 xr T er = r =1 ∞ ar x r = r =1 ar x r . r =1 (iii) If x ∈ c00 , then we can find an n such that xr = 0 for r ≥ n. Thus ∞ n−1 Ta x = ar x r ar x r r =1 r =1 is well defined. If x, y ∈ c00 and λ, µ ∈ R we can find an n such that xr = yr = 0 for r ≥ n. Thus n−1 T a(λx+µ )= n−1 ar (λxr +µyr ) = λ r =1 n−1 ar xr +µ r =1 ar yr = λTa x+µTa y. r =1 We have shown that Ta ∈ c00 . If x ∈ c00 and x = 0 then Tx x = 0. Thus c′00 separates c00 . (iv) Let θ : s → c′00 be given by θa = Ta . By (iii), θ is well defined and, by (ii), θ is surjective. If x ∈ c00 , then we can find an n such that xr = 0 for all r ≥ n. n Tλa+µb x = n xr (λar + µbr ) = λ r =1 n x r ar + µ r =1 xr br r =1 = λTa x + µTb x = (λTa + µTb )x Thus Tλa+µb = λTa + µTb and θ is linear. Further, Ta = 0 ⇔ Ta ej = 0 ∀j ⇔ aj = 0 ∀j ⇔ a = 0, so θ is injective. Thus c′00 is isomorphic to s. 274 Exercise 11.3.10 (i) If x ∈ c00 , then we can find an n such that xr = 0 for r ≥ n. We have n−1 x= xr er r =1 (ii) Let If wj ∈ Rn+1 be the row vector whose k th entry is the m + k th entry of fj . Since n vectors cannot span a space of dimension n + 1 there exists a vector u ∈ Rn+1 with Set bm+r = ur . u ∈ span{w1 , w2 , . . . , wn+1 } / n−1 (iii) Let m(1) = 0 and let m(n) = r=1 (r + 1) for n ≥ 2. For each n ≥ 1 choose ar with m(n) + 1 ≤ r ≤ m(n + 1) so that n λ j fj = a j =1 has no solution with λj ∈ R [1 ≤ j ≤ n]. By construction, a does not lie in the span of the fj . Thus s cannot be spanned by a countable set and is not isomorphic to c00 which can. Thus c00 which is isomorphic to s is not isomorphic to c′00 . 275 Exercise 11.4.2 We use the fact that ej (xk ) = δjk repeatedly. To see that the ej are linearly independent, observe that n j =0 n λj ej = 0 ⇒ j =0 λj ej (xk ) = 0 ∀k ⇒ λk = 0 ∀k. To see that the ej span, observe that, if P ∈ Pn , then n P− P (xj )ej j =0 is a polynomial of degree at most n which vanishes at the n + 1 points xk and is this identically zero. Thus n P= P (xj )ej . j =0 We have ej (xk ) = δjk = ek ej so ˆ n λj ej ek ˆ j =0 and ek P = P (xk ). ˆ n = n λj ek ej = ˆ j =0 λj ej (xk ) j =0 276 Exercise 11.4.4 Observe that ˆ ˆ ej ek = ek ej = δjk ˆ ˆ for all k so ej = ej . 277 Exercise 11.4.5 We have f1 = a e 1 + b e 2 , f2 = ce1 + de2 ˆ and, since the ej form a basis, ˆ1 = Ae1 + B e2 , ˆ ˆ f ˆ2 = B e1 + De2 ˆ ˆ f Now ˆ ˆ 1 = ˆ1 f1 = (Ae1 + B e2 )(ae1 + be2 ) = Aa + Bb f and ˆ ˆ 0 = ˆ1 f2 = (Ae1 + B e2 )(ce1 + de2 ) = Ac + Bd. f Similarly Ca + Db = 0, Cc + Dd = 0. In other words ac bd AB CD so Q= and AB CD = ac bd =I −1 = (ad − bc)−1 a −c . −b d ˆ1 = (ad − bc)−1 ae1 − b(ad − bc)−1 e2 , ˆ ˆ f ˆ2 = −c(ad − bc)−1 ae1 + (ad − bc)−1 de2 . ˆ ˆ f 278 Exercise 11.4.6 Using the summation convention, ˆ δjs = ˆs fj = krs lij er ei = krs lij δir = krs lrj f Thus I = KLT and K = (LT )−1 . 279 Exercise 11.4.9 Let A and B be n × n matrices and α and β the corresponding linear maps for some fixed basis. We know that (α + β )′ = α′ + β ′ , (λα)′ = λα′ , α′′ = α (since we work in a finite dimensional space) ι′ = ι, (αβ )′ = β ′ α′ so (A+B )T = AT +B T , (λA)T = λAT , AT T = A, I T = I, (AB )T = B T AT . If A is invertible, then α is, so α′ is with (α′ )−1 = (α−1 )′ , so AT is invertible with (AT )−1 = (A−1 )T . 280 Exercise 11.4.10 Let α have matrix A with respect to some basis. Then α′ has matrix AT with respect to the dual basis so (by the matrix result) det α′ = det AT = det A = det α. It follows that det(tι − α) = det(tι − α)′ = det(tι − α) and α and α′ have the same characteristic polynomials and so the same eigenvalues. The trace is minus the coefficient of tn−1 in the characteristic polynomial (supposed of degree n). Thus Tr α′ = Tr α. 281 Exercise 11.4.12 We certainly have the zero map 0 ∈ W 0 . If λ, µ ∈ F u′ , v′ ∈ W 0 ⇒ (λu′ + µv′ )w = λu′ w + µv′ w = λ0 + µ0 = 0 ∀w ∈ W ⇒ λu′ + µv′ ∈ W 0 . Thus W 0 is a subspace of U ′ . 282 Exercise 11.4.15 n n j =1 xj e j ∈ W 00 ⇔ xj e j j =1 r =k+1 yr er = 0 ∀yr ∈ F n ⇔ xj e j j =1 er = 0 ∀k + 1 ≤ r ≤ n ⇔ xr = 0 ∀k + 1 ≤ r ≤ n ⇔ j =1 xj e j ∈ W 283 Exercise 11.4.19 dim{u ∈ U : αu = λu} = dim(λι − α)−1 0 = dim im(λι − α) = dim im(λι − α)′ = dim im(λι − α′ ) = dim{u′ ∈ U ′ : α′ u = λu′ .} The eigenvalues and the dimensions of the spaces spanned by the corresponding eigenvectors are the same for α and α′ . 284 Exercise 11.4.20 v′ ∈ (αU )0 ⇒ v′ (αu) = 0 ∀u ∈ U ⇒ (α′ v′ )u = v′ (αu) = 0 ∀u ∈ U ⇒ α′ v ′ ∈ U 0 so α′ (αU 0 ) is a subspace of U . (Remark that (αU )0 is a subspace of V ′ so α′ (αU )0 is a subspace of U ′ .) ˆˆ Let V have basis e1 , e2 and V ′ dual basis e1 e2 . Let α be the endomorphism given by α(e1 ) = e1 , α(e2 ) = 0. We observe that ˆ α′ (ˆ1 ) = e1 , α′ˆe2 ) = 0. e ( ˆ If U = span{e1 }, then αU = U , (αU )0 = span{e2 } and α′ (αU )0 = {0} = U 0 . If If W = span{e2 }, then αW = {0}, (αW )0 = V and ˆ α′ (αU )0 = span{e1 } = U 0 285 Exercise 12.1.2 If U = U1 ⊕ U2 ⊕ . . . ⊕ Um and u ∈ U then, by Definition 12.1.1 (i), we can find uj ∈ Uj such that u = u1 + u2 + . . . + um . If vj ∈ Uj and u = v1 + v2 + . . . + v m , then uj − vj ∈ Uj and 0 = u − v = (u1 − v1 ) + (u2 − v2 ) + . . . + (um − vm ) so, by Definition 12.1.1 (ii), uj − vj = 0 and uj = vj ∀j . If the equation u = u1 + u2 + . . . + um has exactly one solution with uj ∈ U , then the conditions of Definition 12.1.1 can be read off. 286 Exercise 12.1.3 (i) If u ∈ U then we can find uj ∈ Uj such that u = u1 + u2 + . . . + um . For each j we can find λjk ∈ F such that n(j ) λjk ejk uj = k=1 and so m n(j ) u= λjk ejk . j =1 k=1 Thus the ejk span U . If m n(j ) 0= λjk ejk j =1 k=1 n(j ) k=1 λjk ejk we have uj ∈ Uj and, applying the then setting uj = definition of direct sum, we obtain n(j ) λjk ejk , 0 = uj = k=1 so λjk = 0 for all k and j . Thus the ejk are linearly independent and so form a basis. (ii) and (iii) Since every subspace of a finite dimension subspace is finite dimensional U finite dimensional ⇒ Uj finite dimensional If the Uj are finite dimensional the we can choose bases as in (i) and observe that U is finite dimensional with m dim U = m n(j ) = j =1 dim Uj . j =1 287 Exercise 12.1.4 Observe that B = U ∩ V is a subspace of a finite dimensional space so has a basis e1 , e2 , . . . , ek . Since U is a subspace of a finite dimensional space containing B we can extend the basis of B to a basis of U e1 , e2 , . . . , ek , ek+1 , ek+2 , . . . , ek+l . Similarly we can extend the basis of B to a basis of v e1 , e2 , . . . , ek , ek+l+1 , ek+l+2 , . . . , ek+l+r . Let C = span{ek+1 , ek+2 , . . . , ek+l , ek+l+1 , ek+l+2 , . . . , ek+l+r }. We show that the ej are linearly independent as follows. If k +l +r λj ej = 0 j =1 then k +l j =1 k +l +r λj ej = − Thus j =k+l+1 k +l λj ej ∈ U ∩ V = B k λj ej = j =1 µj ej j =1 for some µj . Thus λj = 0 for 1 ≤ j ≤ k . Thus k k +l +r λj ej + j =1 λj ej = 0 j =k+l+1 so λj = 0 for k + l + 1 ≤ j ≤ k + l + r and for 1 ≤ j ≤ k . We can now read off the desired results, U = A ⊕ B, W = B ⊕ C and U + W = U ⊕ C = W ⊕ A. Observe that if U = A′ ⊕ B ′ , W = B ′ ⊕ C ′ , U + W = U ⊕ C ′ = W ⊕ A ′ we have B ′ ⊆ U, V so B ′ ⊆ U ∩ V. If v ∈ U ∩ W then we can find b1 , b2 ∈ B ′ and a ∈ A, c ∈ C such that v = a + b1 = c + b2 so Thus a − c = b1 − b2 ∈ B ′ . a = c + (b1 − b2 ) 288 with a ∈ A′ ⊆ U and b1 − b2 ∈ U ∩ W ⊆ U . Since U + W = U ⊕ C ′ we have c = 0. so v = c + b2 = b2 ∈ B ′ . Thus B′ = U ∩ W and we have shown B unique. Not unique. Take V = R3 (using row vectors) U = {(x, y, 0) : x, y ∈ R}, V = {(x, 0, z ) : x, z ∈ R} B = {(x, 0, 0) : x ∈ R}, A = {(0, y, 0) : y ∈ R}, A′ = {(y, y, 0) : y ∈ R} Then and C = {(0, 0, z ) : y ∈ R}, C ′ = {(z, 0, z ) : z ∈ R}. U = A ⊕ B, W = B ⊕ C and U + W = U ⊕ C = W ⊕ A U = A′ ⊕ B, W = B ′ ⊕ C and U + W = U ⊕ C ′ = W ⊕ A′ . 289 Exercise 12.1.6 (i) Can occur. Take V = V1 = V2 = F2 . (ii) Cannot occur. If e1 , e2 , . . . ek is a basis for V1 and ek+1 , ek+2 , . . . ek+m is a basis for V2 then e1 , e2 , . . . ek+m span V1 + V2 so dim(V1 + V2 ) ≤ k + m = dim V1 + dim V2 (ii) Can occur. Let e1 , e2 be a basis for V = F2 and V1 = V2 = span{e1 }. 290 Exercise 12.1.7 We know from Theorem 11.2.2 that there exists an m such that α U = αm U . We thus have αm U an invariant subspace. m+1 If W is an invariant subspace W = αm V ⊆ αm U. Thus αm U is the unique maximal invariant subspace. (ii) If x ∈ M ∩ N then αm x = 0 But α|M : M → M is surjective so invertible so α|m is so x = 0. By M the rank-nullity theorem dim M + dim N = dim U so U + M ⊕ V . (iii) α(M ) = M ⊆ M , αm α(N ) = ααm N = 0 so αN ⊆ N . (iv) As we observed in (ii), β : M → M is surjective, so β is invertible, so an isomorphism. We know that γ m = 0. (v) Observe that M ′ is an invariant subspace so M ′ ⊆ M . If b ∈ N ′ then αr (b) = (γ ′ )r b = 0 for r large so b ∈ N . Thus N ′ ⊆ N . We have dim M ′ ≤ dim M , dim N ′ ≤ dim N dim U = dim M ′ + dim N ′ ≤ dim M + dim N = dim U so dim M ′ = dim M , dim N ′ = dim N and M = M ′ , N = N ′ . By considering b = 0, we have β = β ′ and similarly, by considering a = 0, we have γ = γ ′ . (vi) Let A correspond to a linear map α for some choice of basis for a space U of appropriate dimension. Choose a basis M for M defined as above and a basis N for N . Then E , N is a basis for U with respect to which α has the form B0 0C with B an invertible r × r matrix and C a nilpotent n − r × n − r matrix, The result now follows from the change of basis formula. 291 Exercise 12.1.11 Let dim U = n and dim V = r. If U is neither V nor {0}, then 0 < r < n. Choose a basis e1 , e2 , . . . , e r and extend it to a basis e1 , e2 , . . . , e n of V . If W = span{er+1 , e2 , . . . , en } W ′ = span{e1 + er+1 , e2 , . . . , en }, then V = U ⊕ W = U ⊕ W ′ but er+1 ∈ W \ W ′ so W = W . 292 Exercise 12.1.13 (i) It will remain true θ has an n2 × n2 diagonal matrix with dim U diagonal entries taking the value 1 and dim V diagonal entries taking the value −1, since θ|U = ι|U and since θ|V = −ι|V . (ii) det θ = (−1)dim V = (−1)n(n−1)/2 . Now n2 + 8n + 16 − n − 4 − n2 + n (n + 4)((n + 4) − 1) n(n − 1) − = = 4n+6 2 2 2 is divisible by 2 so det θ depends only on the value of n modulo 4. By inspection n ≡ 0 ⇒ det θ n ≡ 1 ⇒ det θ n ≡ 2 ⇒ det θ n ≡ 3 ⇒ det θ = 1, = 1, = −1, = −1. 293 Exercise 12.1.14 Automatically 0 ∈ V and, if λ, µ ∈ R, f, g ∈ V ⇒ (λf + µg )(−x) = λf (−x) + µg (−x) = −λf (x) − µg (x) = −(λf + µg )(x) ∀x ⇒ λf + µg ∈ V, so V is subspace. A similar argument shows that U is a subspace, If f ∈ C (R), then u(x) = (f (x)+ f (−x))/2, v (x) = (f (x) − f (−x))/2 define u ∈ U , v ∈ V . Since f = u + v we have U + V = C (R). Now if f ∈ U ∩ V , then f (x) = f (−x) = −f (x) so f (x) = 0 for all x ∈ R and f = 0. Thus U ∩ V = {0} and U and V are complementary subspaces. 294 Exercise 12.1.15 (i)⇒(ii) Let U = αV and W = (ι − α)V . If u ∈ U , then u = αv for some v ∈ V so αu = α2 v = αv = u so α|U = ι|U . If u ∈ W , then u = (ι − α)v for some v ∈ V , so so α|W = 0|W αu = (α − α2 )v = αv − αv = 0, Since v = αv + (ι − α)v for all v ∈ V so V = U + W . If u ∈ U ∩ W , then u = α|U u = αu = α|W u = 0. Thus U ∩ W = {0} and V = U ⊕ W . (ii)⇒(iii) Let e1 , e2 , . . . , er be a basis of U and er+1 , . . . , en be a basis of W . Then with respect to the basis e1 , e2 , . . . , en of V , α has a matrix of stated type. (iii)⇒(i) Since A2 = A we have α2 = α. 2 We have α1 (x, y ) = (x, 0) = α1 (x, y ) so α1 is a projection. 2 We have α2 (x, y ) = (0, y ) = (0, 0) = α2 (x, y ) if y = 1, so α2 is not a projection. 2 We have α3 (x, y ) = (x, y ) = (y, x) = α3 (x, y ) if x = 0, y = 1, so α3 is not a projection. 2 We have α4 (x, y ) = (x + y, 0) = (x + y, 0) = α4 (x, y ), so α4 is a projection. 2 We have α5 (x, y ) = (x + y, x + y ) = 2(x + y ), 2(x + y ) = α5 (x, y ) if x = 0, y = 1, so α5 is not a projection. We have α6 (x, y ) = tion. 1 (x + y ), 1 (x + y ) 2 2 2 = α6 (x, y ) so α6 is a projec- 295 Exercise 12.1.16 If α is a projection, then (ι − α)2 = ι − 2α + α2 = ι − 2α + α = ι − α so ι − α is a projection. If ι − α is a projection, then the first paragraph tells us that α = ι − (ι − α) is a projection. 296 Exercise 12.1.17 If α and β are projections and αβ = βα, then (αβ )2 = (αβ )(αβ ) = α(βα)β = α(αβ )β = α2 β 2 = αβ. We work in R2 with row vectors. If β (x, y ) = (x + y, 0), α(x, y ) = (0, y ) then α2 = α, β 2 = α βα(x, y ) = (y, 0), αβ (x, y ) = (0, y ) 2 so (αβ ) = (αβ ) but (βα)2 = (βα). If β (x, y ) = (x + y, 0), α(x, y ) = (0, x + y ) then α = α, β = α 2 2 βα(x, y ) = (x + y, 0), αβ (x, y ) = (0, x + y ) 2 so (αβ ) = (αβ ) and (βα)2 = (βα), but (looking e.g. at (x, y ) = (1, 1)) αβ = αβ . 297 Exercise 12.1.18 (i) If αβ = −βα, then αβ = ααβ = α(−βα) = (−αβ )α = βαα = βα Thus αβ = 0 and βα = 0. (ii) If (α + β ) is a projection, then α + αβ + βα + β = (α + β )2 = α + β, so αβ = −βα and αβ = βα = 0. If αβ = βα = 0, then (α + β )2 = α + αβ + βα + β = α + β, so α + β is a projection. (ii) If (α − β ) is a projection, then α − αβ − βα + β = (α − β )2 = α − β, so −αβ − βα = 2β whence (ι − α)β = −β (ι − α) so, since ι − α is a projection, (ι − α)β = β (ι − α) = 0 and αβ = βα = β . If αβ = βα = β , then (α − β )2 = α − αβ − βα + β = α − β, so α − β is a projection. 298 Exercise 12.1.19 If α is diagonalisable with distinct eigenvalues λ1 , λ2 , . . . , λm , then U is the direct sum of the spaces Ej = {e : αe = λj e}. If u ∈ U , we can write u uniquely as u = e1 + e2 + . . . + e m with ej ∈ Ej If we set πj (e) = ej for 1 ≤ j ≤ m then direct verification shows that πj is linear and πj is a projection. By inspection πk πj = 0 when k = j and α = λ1 π 1 + λ2 π 2 + . . . + λ m π m . Conversely, if the stated conditions hold and we write Ej = πj U , we see that Ej is a subspace and πi Ej = πi πj Ej = 0 for i = j . Since iota = π1 + π2 + . . . + πm we have m u = ιu = m πj u = j =1 uj j =1 with uj = πj u ∈ Ej . On the other hand, if ej ∈ Ej and m ej = 0 j =1 then applying πi to both sides we get ei = 0 for all i. Thus U = E 1 ⊕ E2 ⊕ . . . ⊕ E m . Let Ej be a basis for Ej . The set E = conditions πk πj = 0 when k = j and m j =1 Ej is a basis for U . The α = λ1 π 1 + λ2 π 2 + . . . + λ m π m show that E is a basis of eigenvectors so α is diagonalisable. 299 Exercise 12.2.1 1≤j ≤n Let E (rs) = (δir δjs )1≤i≤n Then, if A = (aij ), n n A= ars E (rs) r =1 s=1 so the E (r, s) span Mn (F). Also n n r =1 s=1 brs E (rs) = 0 ⇒ B = 0 ⇒ brs = 0 ∀r ∀s so the E (r, s) form a basis for Mn (F). Thus dim Mn (F) = n2 . It follows that the n2 + 1 elements Aj with 0 ≤ j ≤ n2 must be linearly dependent, ie we can find aj ∈ C, not all zero, such that n2 j j =0 aj A = 0. In other words, there is a non-trivial polynomial P of degree at most n2 such that P (A) = 0. 300 Exercise 12.2.2 (i) Observe that n χD (t) = j =1 (t − λj ). Let Dj be the diagonal matrix with the same entries as D except that the j th diagonal entry is 0. Then n n bk D k = k=0 j =1 n (D − λj I ) = Dj = 0. j =1 (ii) We have A = P −1 DP for some invertible matrix P . χD = χA and χA (A) = χD (A) = χD (P −1 DP ) = P −1 χD (D)P = P −1 0P = 0 as stated. 301 Exercise 12.2.3 (i) Let C = B −1 . Using the summation convection, Tr B −1 AB = Tr CAB = cij ajk bki = bki cij ajk = δkj ajk = akk = Tr A. (Lots of other proofs available.) (ii) QA (tI − A) = tδii − aii = nt − Tr A. Part (i) yields QB −1 AB = QA . (iii) QA (A) = nA − (Tr A)I . If a11 = −a22 = 1 and aij = 0 otherwise, the. QA (A) = 0. If n = 1, Q(a11 ) (t) = t − a11 and Q(a11 ) (a11 ) = (a11 ) − a11 (1) = (0). 302 Exercise 12.2.6 The roots of the characteristic polynomial of a triangular matrix are the diagonal entries. If every real n × n was triangularisable by change of basis, then, since the characteristic equation is unaltered by change of basis, every root of the characteristic equation would be real. Taking A = (aij ) with arr = 1 for r = 1, 2 a12 = −a21 = 1, aij = 0, otherwise, we see that this is not the case. If dim V = 1 every matrix is triangular. 303 Exercise 12.2.7 (i) Choose a basis such that α has an upper triangular matrix with respect to that basis. The eigenvalues of α are the diagonal entries. With respect to the chosen basis αr has matrix Ar which is upper triangular with diagonal entries the rth powers of the diagonal entries of A. Thus αr has an eigenvalue µ if and only if α has an eigenvalue λ with λr = µ. (ii) If α is invertible then αr has an eigenvalue µ if and only if α has an eigenvalue λ with λr = µ. To prove this observe that if β is invertible with eigenvalue λ and associated eigenvector e β −1 β e = e so λ = 0 and β −1 has eigenvalue λ−1 and associated eigenvector e. (iii) False. Let 0 −1 . 10 Then A has no real eigenvalues but A2 = I has 1 as eigenvalue. 0 a12 a13 b11 b12 b13 000 c11 c12 c13 0 a22 a23 0 0 b23 0 c22 c23 = 0 0 0 . 0 0 a33 0 0 b33 000 0 0 0 A= 304 Exercise 12.2.8 (i) We have b11 b12 0 a12 a13 0 a22 a23 0 0 0 0 0 0 a33 0 a12 a13 b11 c11 0 a22 a23 0 = 0 0 a33 0 000 0 0 0 = 000 000 c11 c12 c13 b13 b23 0 c22 c23 = 0 0 0 000 0 0 0 b33 b11 c12 + b12 c22 b11 c13 + b12 c23 0 0 0 0 (ii) Let Tj be an n × n upper triangular matrix with the j th diagonal entry 0. Then T1 T2 . . . Tn = 0 We can prove this by induction on n using matrices or as follows. Let τj be the linear map on an n dimensional vector space U which has matrix Tj with respect to some basis e1 , e2 , . . . , en . Then writing Ej = span{e1 , . . . , ej } for 1 ≤ j ≤ n and E0 = {0} we have τj Ej ⊆ Ej −1 . Thus τ1 τ2 . . . τn U = τ1 τ2 . . . τn En = E0 and τ1 τ2 . . . τn = 0 so T1 T2 . . . Tn = 0 (iii) Not necessarily true. 11 111 0 1 1 0 0 000 00 1 0 = 0 1 0 = 0 011 0 1 0 1 1 0 1 1 1 001 11 111 1 0 0 1 1 11 00 001 1 5 11 3 1 = 2 11 1 00 1 so the matrix product is not zero. 305 Exercise 12.2.9 By induction on r, βαr = αr β , so, by induction on s, β s αr = αr β s . Thus n m aj α j =0 j n m k m j ak β = b k aj β k α j aj b k α β = j =0 k=0 m k=0 n k k=0 j =0 n ak β k = k=0 aj α j . j =0 306 Exercise 12.2.11 Setting t = 0 we have 0 = det A = a0 By the Cayley–Hamilton theorem n aj Aj a0 I = j =1 so, multiplying by a−1 A−1 , 0 we get n −1 A = −a−1 0 aj Aj −1 . j =1 Does not appear to be a good method. The computation of the aj appears to require the evaluation of det(tI − A) which appears to require at least as many operations as inverting a matrix by Gaussian elimination. 307 Exercise 12.2.12 Let M be the smallest value of m with αm = 0 We know that there exist aj such that n−1 n aj α j α= j =0 n and so either α = 0 or there exists a k with n − 1 ≥ k ≥ 0 such that n−1 n aj α j α= j =k and ak = 0. But then applying α M −k −1 to both sides we get 0 = ak α M − 1 so αM1 = 0, contrary to our definition of M . 308 Exercise 12.3.1 (i) We have det(tI − A1 ) = det(tI − A2 ) = t2 . A1 and A2 are not similar since they have different ranks. (ii) We have det(tI − A1 ) = det(tI − A2 ) = det(tI − A3 ) = t3 . None of A1 , A2 , A3 are similar since they have different ranks. (iii) Set and 0 0 A8 = 0 0 0 0 0 0 0 0 0 0 A10 Then 0 0 0 0 , A9 = 0 0 0 0 0 0 = 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 . 0 0 0 0 1 0 det(tI − Aj ) = t4 for 6 ≤ j ≤ 10. Now A6 has rank 1, A7 and A9 have rank 2, A8 has rank 0 and A9 has rank 3. 0 0 A2 = 0, A2 = 7 9 0 0 2 2 so A7 has rank 0 and A9 rank 1. None 0 0 0 0 of 1 0 0 0 the 0 0 0 0 Aj can be similar. 309 Exercise 12.3.3 (i) Since the characteristic polynomials have the form tn so do the minimal polynomials. The minimal polynomial of Ak is mk (t) = tr r where Ar = 0, Ak−1 = 0. Thus k m1 (t) = t, m2 (t) = t2 , m3 (t) = t, m4 (t) = t2 , m5 (t) = t3 m6 (t) = t2 , m7 (t) = t2 , m8 (t) = t, m (ii) We have det(tI − A) = (t − 1)2 (t − 2)2 = det(tI − B ) Now (A − I )(A − 2I )2 = 0 and (A − I )2 (A − I )2 = 0 so A has minimal polynomial (t − 1)2 (t − 2)2 . We also have (B − I )(B − 2I ) = 0 and (B − II )2 = 0 but (B − I ) (B − 2I ) = 0 so A has minimal polynomial (t − 1)2 (t − 2). 2 310 Exercise 12.3.6 (i) r=1 (D − λi I ) i r (djj − λi ) = 0 so i=1 is a diagonal matrix with j th diagonal entry r i=1 (D − λi I ) = 0. Now suppose, without loss of generality that d11 = λ1 . Then r=2 (D− i λi I ) is a diagonal matrix with 1st diagonal entry r=2 (d11 − λi ) = 0 so i r r i=2 (D − λi I ) = 0. Thus i=1 (t − λi ) is the minimal polynomial for D. (ii) Choose a basis of eigenvectors. e1 , e2 , . . . en . With respect to this basis α has a diagonal matrix with j th diagonal entry the eigenvalue associated with ej . Thus if λ is associated with k of the basis vectors det(tI − D) = det(tι − α) has (t − λ)k as a factor but not (t − λ)k+1 Now writing S for the set of eigenvalues which are not equal to λ dim Uλ = dim( µ ∈S (µι−α)U ) = dim span{ µ ∈S (µι−α)U )ej : 1 ≤ j ≤ n} = k 311 Exercise 12.3.7 (i) If P has a repeated root θ, then P (t) = (t − θ)k Q(t) with Q a polynomial and k ≥ 2. We have P ′ (t) = k (t − θ)k−1 Q(t) + (t − θ)k Q′ (t) = (t − θ)k−1 (kQ(t) + (t − θ)Q′ (t) so t − θ is a factor of both P (t) and P ′ (t). (ii) Since Am = I the minimal polynomial mA of A must divide tm −1. But tm − 1 = 0 has no repeated roots (by part (i), or otherwise). Thus ma has no repeated roots so A is diagonalisable. If A is a real symmetric matrix then its eigenvalues are real so we have A = LDL−1 with L invertible (indeed we may take L orthogonal) and D real diagonal Dm = I so dm = 1 and djj is real so djj = ±1 so D2 + I so A2 = I . jj 312 Exercise 12.3.9 (i) Let ∂Q denote the degree of Q. If P = {0} then the set E = {∂P : P ∈ P \ {0}} is a non-empty subset of the positive integers and so has a least member N. Let Q0 ∈ P have degree N and leading coefficient a. If we set P0 = a−1 Q0 , then P0 is a monic polynomial of smallest degree in P . If Q ∈ P then Q = SP + R with S , R polynomials and ∂R < N . Now R = Q − SP0 ∈ P so by the definition of N , R = 0 and Q = SP0 . (ii) Observe that P satisfies the hypotheses of (i). If S divides each Pj then S divides each Qj Pj and so divides P0 . (iii) Take Pj (t) = r j =1 Qj (t) i=j (t − λi )m(i) and apply (ii). 313 Exercise 12.3.11 Write α = α1 ⊕ α2 ⊕ . . . ⊕ αr The expansion u = u1 + u2 + . . . + ur with uj ∈ Uj always exists a and is unique so α(u) == α1 u1 + α2 u2 + . . . + αr ur is well defined. If u, v ∈ U and λ, µ then we can write u = u1 + u2 + . . . + ur v = v 1 + v2 + . . . + v r with uj , vj ∈ Uj . We then have λu + µv = (λu1 + µv1 ) + (λu2 + µv2 ) + . . . + (λur + µvr ) and λuj + µvj ∈ Uj so α(λu + µv) = α1 (λu1 + µv1 ) + α2 (λu2 + µv2 ) + . . . + αr (λur + µvr ) = (λα1 u1 + µα1 v1 ) + (λα2 u2 + µα2 v2 ) + . . . + (λαr u1 + µαr v1 ) = λ(α1 u1 + α2 u2 + . . . + αr ur ) + µ(α1 v1 + α2 v2 + . . . + αr vr ) = λαu + µαv so α is linear. 314 Exercise 12.3.13 (i) We have r ι= Qj (α) j =1 i=j (α − λj ι)m(j ) , so, if u ∈ U we have r u= Qj (α) j =1 i=j (α − λj ι)m(j ) u = sumr=1 uj j where uj = Qj (α) i=j (α − λj ι)m(j ) u. Since (α − λj ι)m(j ) uj = (α − λj ι)m(j ) Qj (α) i=j (α − λj ι)m(j ) u r = Qj (α) j =1 (α − λj ι)m(j ) u = Qj (α)Q(α)u = Qj (α)0 = 0 we have uj ∈ Uj . Thus U = U1 + U2 + . . . + Ur . (ii) Next we observe that if v ∈ Uk with k = j then Qj (α) i=j (α − λj ι)m(j ) v = 0 Thus if v ∈ Uk we have r v= Qj (α) j =1 In particular i=j i=k (α (α − λi ι)m(i) v = Qk (α) i=k (α − λi ι)m(i) v. − λi ι)m(i) v = 0 if v = 0. Now if vj ∈ Uj and r vj = 0 j =1 then r m(i) i=k (α − λj ι) m(i) vk = i=k (α − λj ι) so vk = 0 for each k . Thus U = U1 ⊕ U2 ⊕ · · · ⊕ Ur vj 0 j =1 315 (iii) We have u ∈ Uj ⇒ (α−λj )m(j ) u = 0 ⇒ (α−λj )m(j ) αu = α(α−λj )m(j ) u = 0 ⇒ αu ∈ Uj . If u ∈ U then r u= uj j =1 and r αu = α r r uj = αuj = j =1 j =1 j =1 αj uj = α = α1 ⊕ α2 ⊕ · · · ⊕ αr u so α = α1 ⊕ α2 ⊕ · · · ⊕ αr . (iiv) By the definition of Uj , (α − λj ι)m(j ) Uj = 0 so the minimal polynomial for αj must have the form. (t − λj )p(j ) . But If u ∈ U then u = r=1 us with us ∈ Us s r r p( j ) j =1 (α − λj ) r r p( j ) u= s=1 j =1 (α − λj ) so the minimal polynomial must divide for all J . uj = 0 = 0. s=1 r p( j ) j =1 (α − λj ) so p(j ) = m(j ) 316 Exercise 12.3.14 Let e1 , e2 , . . . , er be a basis for U with respect to which α has matrix A and er+1 , er+2 , . . . , er+s be a basis for V with respect to which β has matrix B Then e1 , e2 , . . . , er+s is a basis for W with respect to which α ⊕ β has matrix A0 0B Thus tI − A 0 = det(tI −A) det(tI −B ) = χα (t)χβ (t). χα⊕β (t) = det 0 tI − B If α has minimal polynomial mα , β has minimal polynomial mβ and P is a polynomial P (α ⊕ β ) = 0 ⇔ P (α ⊕ β )(u + v) = 0∀u ∈ U, v ∈ V ⇔ P (α)u + P (β )v = 0∀u ∈ U, v ∈ V ⇔ P (α)u = P (β )v = 0∀u ∈ U, v ∈ V ⇔ mα and mβ divide P Thus the minimal polynomial of α ⊕ β is the lowest common multiple of (ie the monic polynomial of lowest degree dividing) mα and mβ . 317 Exercise 12.3.15 (i) If (t − λ) is a factor of the characteristic polynomial then λ is an eigenvalue and (t − λ) must be a factor of the minimal polynomial. Thus we cannot choose S (t) = t + 1, Q(t) = t. (ii) If m = 1 set A = 0. If m = n let A = (aij ) be the n × n matrix with ai,i+1 = 1 aij = 0 otherwise. If n > m > 1 let B = 0n−m the n − m + ×n − m zero matrix and let C be the m × m matrix with ci,i+1 = 1 cij = 0 otherwise. Take A= B0 0C A − λI has characteristic equation (t − λ)n and minimal polynomial (t − λ)m (iii) We can write P (t) = i = 1r (t−λi )m(i) and Q(t) = i = 1r (t− λi )m(i) with the λj distinct and m(j ) ≥ p(j ) ≥ 1. By (ii) we can find a m(j ) × m(j ) matrix Bj = Aj − λIm(j ) with characteristic polynomial (t − λ)m(j ) and minimal polynomial (t − λ)p(j ) . Take A1 0 . . . 0 0 0 A2 0 . . . 0 0 . . . . . . . A= . . . . . 0 0 0 ... A 0 r −1 0 0 0 ... 0 Ar A has the required properties either by direct computation or using Exercise 12.3.14. 318 Exercise 12.4.3 Since αej = ej +1 for 1 ≤ j ≤ n − 1 and αen = 0 we have A = (aij ) with ai,i+1 = 1 aij = 0 otherwise. 319 Exercise 12.4.4 If αn = 0, then e, αe, . . . αn e are linearly independent a and a space of dimension n would contain n + 1 linearly independent vectors which is impossible. 320 Exercise 12.4.7 We have αk e = 0 for only finitely many values of k . 321 Exercise 12.4.11 Observe that (djj ) is a 1 × 1 Jordan matrix. 322 Exercise 12.4.12 (i) A is the matrix of a linear map α : U → U α = α1 ⊕ α2 . . . αr with U = U1 ⊕ U2 . . . Ur where αj is a linear map on Uj with αj = k λj ι + βj with βj a nilpotent linear map on Uj with βj j = 0, β kj −1 = 0 dim Uj = kj Thus r (λι − α)−k 0 = j =1 (λι − αj )−k 0 and r dim(λι − α)−k 0 = j =1 dim(λι − αj )−k 0 = λ=λj min{k, kj }. Thus {x ∈ Cn : (λI − A)k x = 0} = λ=λj min{k, kj } (ii) By (i) λ=λj min{k, kj } = ˜ λ=λi ˜ min{k, ki } ˜ ˜ so r = r and, possibly after renumbering, λj = λj and kj = kj for ˜ 1 ≤ j ≤ r. 323 Exercise 12.4.13 Observe by looking at the effect on z n that T n+1 = 0 but T n = 0. Thus, since Pn has dimension n + 1, T has Jordan form 0 1 0 0 ... 0 0 0 0 1 0 . . . 0 0 0 0 0 1 . . . 0 0 Jn+1 (0) = . . . . . . . . . . . . . . . . . . . 0 0 0 0 . . . 0 1 0 0 0 0 ... 0 0 We observe that er (z ) = z r is an eigenvector with eigenvalue r [n ≥ r ≥ 0]. Thus the Jordan normal form is the diagonal matrix 0 0 0 0 ... 0 0 0 1 0 0 . . . 0 0 0 0 2 0 . . . 0 0 . . . . .. . . . . . . . . . .... 0 0 0 0 . . . n − 1 0 0 0 0 0 ... 0 n 324 Exercise 12.4.14 (i) If λ is a root of χα then we can find an associated eigenvector e. Since e ∈ dim{u : (α − λι)(u) = 0} we have mg (λ) ≥ 1. Now take a basis e1 , e2 . . . , emg (λ) for {u : (α − λι)(u) = 0} and extend it to a basis e1 , e2 . . . , en for U . With respect to this basis α has matrix λImg (λ) 0 A= R S with Img (λ) the mg (λ) × (mg (λ) identity matrix so and ma (λ) ≥ mg (λ). χα (t) = (t − λ)mg (λ) χR (t) (ii) Suppose s ≥ r ≥ 1 Let B = (bij ) be the r × r matrix with bi,i+1 = 1 and bij = 0 and A(r, s) the s × s matrix with A(r, s) = λB 0 . 00 If αr,s is the linear map associated with A(r, s) then, for this linear map, mg (λ) = r if λ = 0 , = 0 otherwise ma (λ) = s if λ = 0 = 0 otherwise If we now set βr,s = µι − αr,s then for this linear map mg (λ) = r if λ = µ , = 0 otherwise ma (λ) = s if λ = µ = 0 otherwise Now take a vector space U = U1 ⊕ U2 ⊕ . . . ⊕ Ur with dim Uk = na (λk ). By the previous paragraph we can find γk : Uk → Uk linear such that for this linear map mg (λ) = ng (λk )If =0 if λ = λk , otherwise na (λ) = If we set α = γ1 ⊕ γ2 ⊕ . . . ⊕ γ r then the required properties can be read off. s if λ = λk = 0 otherwise 325 (iii) Suppose α has the associated Jordan form Jk1 (λ1 ) Jk2 (λ2 ) Jk3 (λ3 ) A= ... J (λ kr − 1 r −1 ) Jkr (λr ) (using the notation of Theorem 12.4.10). We observe that the characteristic polynomial r r χα (t) = χA (t) = χJkj (t) = j =1 j =1 (t − λj )k(j ) so ma (λ) = k (j ). λj =λ Observe that the dimension of the space of solutions of Jk x = λ x is 1 if λ = λk and zero otherwise we see that ma (λ) = λj =λ 1 = card{ : λj = λ} 326 Exercise 12.5.1 Let U be a finite dimensional vector space over C. If α ∈ L(U, U ) then the Jordan normal form theorem tells that we can write U = j = 1r Uj and find αj ∈ L(Uj , Uj ) such that α = j = 1r αj and Uj has a basis with respect to which αj has an n(j ) × n(j ) matrix λ 1 0 0 ... 0 0 0 λ 1 0 . . . 0 0 0 0 λ 1 . . . 0 0 . . . . Aj = . . . . . .. . . . . .... 0 0 0 0 . . . λ 1 0 0 0 0 ... 0 λ Now r det(tι − α) = and j =1 r det(tι − αj ) r r n(j ) j =1 j =1 (t − λj )n(j ) (α − λj ) n(j ) (since (αj − λj ) = r j=1 i=1 (αj − λi )n(j ) = 0 = 0) so χα (α) = 0. Lemma 12.2.1 tells us that there exists a non-zero polynomial P with P (α) = 0. This is all we need to show that there is a minimal polynomial. The proof of the Jordan form theorem only uses the minimal polynomial. 327 Exercise 12.5.2 Consider the s × s matrix λ 0 0 B = . . . 0 0 1 λ 0 . . . 0 0 0 1 λ . . . 0 0 0 0 1 . . . 0 0 ... 0 0 . . . 0 0 . . . 0 0 . .. . . . . . . . λ 1 ... 0 λ If λ = 0, B is invertible so rank B j = s for all j ≥ 0. If λ = 0 then direct calculation shows that rank B j = max{0, k − j } for all j ≥ 0. If α = r k=1 αk then rank αj = r k=1 j rank αk Thus if the Jordan normal form of an n × n blocks 0 1 0 0 ... 0 0 0 1 0 . . . 0 0 0 0 1 . . . 0 Kp = . . . . . . . . . . . . . . . 0 0 0 0 . . . 0 0 0 0 0 ... 0 with Kp an n(p) × n(p) matrix, then m j rankA = n− A contains m nilpotent 0 0 0 . . . 1 0 m np + p=1 p=1 max{0, np − j } . (ii) Observe that if we set m r(j ) = n− m np p=1 + p=1 max{0, np − j } then for some m = maxp n(p) =≤ n together with the condition r0 − r1 ≥ r1 − r2 ≥ r2 − r3 ≥ . . . ≥ rm−1 − rm . Let q (j ) = (sj − sj −1 ) − (sj +1 − sj ). Let A be the n × n matrix with qj nilpotent Jordan blocks of size j × j and s(m) 1 × 1 Jordan blocks (1). If α has matrix A with respect to some basis then α satisfies the required conditions. 328 Exercise 12.5.3 (i) We have 00 0 0 A1 = 0 0 00 0 0 , 0 0 0 0 A4 = 0 0 0 0 0 0 A2 1 0 0 0 0100 0 0 0 0 = 0 0 0 0 , 0000 0 00 0 1 0 , A5 = 0 0 1 0 00 A3 1 0 0 0 0 0 = 0 0 00 0 0 0 1 00 1 0 0 0 0 1 0 0 0 0 , 0 0 (ii) For A1 xr = 0 so x(t) = c a constant. ˙ For A2 x1 = x2 , xr = 0 otherwise so x(t) = (c2 t + c1 , c2 , c3 , c4 )T with ˙ ˙ cj constants. For A3 x1 = x2 , x2 = x3 , xr = 0 otherwise so x(t) = (c3 t2 /2 + c2 t + ˙ ˙ ˙ T c1 , c2 t + c3 , c3 , c4 ) with cj constants. Similarly for A4 , x(t) = (c4 t3 /6 + c3 t2 /2 + c2 t + c1 , c4 t2 /2 + c3 t + c2 , c4 t + c3 , c4 )T with cj constants. Similarly for A5 x(t) = (c2 t + c1 , c2 , c4 t + c3 , c4 )T (iii) If we write x(t) = eλ y(t) then x′ (t) = Aj x(t) Thus the general solution of y′ (t) = (λI + Aj )y(t) is y(t) = e−λt x(t) where x(t) is the general solution of x′ (t) = Ax(t). (iv) Suppose B1 0 0 B2 . . B= . . . . 0 0 00 ... ... 0 0 . . . . . . Br−1 ... 0 0 0 . . . 0 Br with Bj = λj I + Cj where Cj is an s(j ) × s(j ) nilpotent Jordan matrix. Then the general solution of x′ (t) = B x(t) 329 is x = (y1 , y2 , . . . , yr )T with (s(j )) yj (t) = e−λj t (Pj (s(j )−1) (t), Pj (t), . . . , Pj (t))T with Pj any polynomial of degree s(j ) − 1. (iv) Find M invertible and B a matrix in Jordan form such that M AM −1 = B then writing y = M x we have ⋆ The general solution of ˙ y = B y. x′ (t) = Ax(t). is x = M −1 y where y is the general solution of ⋆. 330 Exercise 12.5.4 Observe that 0 1 0 0 0 1 . . . 0 0 0 0 0 0 −an−1 −an−2 −an−3 ... ... 0 0 x0 x0 0 0 x1 x1 . x2 x2 . . . = . ... 1 0 . . . . ... 0 1 xn−2 xn−2 . . . −a1 −a0 xn−1 xn−1 ⇔ xj = xj +1 [0 ≤ j ≤ n − 2], − ˙ n−2 ⇔ x(n) + aj x(n−j ) = 0 j =1 n−2 aj xj = xn−1 j =0 331 Exercise 12.5.5 Suppose U = k (j )− 1 βj = 0, k βj j r j =1 Uj , Uj has dimension kj βj : Uj → Uj has = 0, αj = λj ι + βj and α = r j =1 αj . Write s(λ) = maxλ=λj kj . If P is a polynomial P (α) = 0 ⇔ P (αj ) = 0 ∀j ↔ (t − λ)s (λ)|P (t). Thus the characteristic polynomial equals the minimal polynomial if and only if r j =1 (t − λ)kj = s(λ)=0 (t − λ)s(λ) that is to say, if and only if all the λj are distinct. 332 Exercise 12.5.6 Writing ej for the vector with 1 in the j th place and zero everywhere else. By induction Aj en = en−j + fn−j where fn−j ∈ span{er : n ≥ r ≥ n − j + 1}. j Thus the A en with n − 1 ≥ j ≥ 0 are linearly independent and n−1 bj Aj j =0 e = 0 ⇔ bj = 0 ∀0 ≤ j ≤ n − 1. Thus the minimal polynomial has degree at least n. Since the minimal polynomial divides the characteristic polynomial and the characteristic polynomial has degree n it follows that the two polynomials are equal. By Exercise 12.5.5, it follows that that there is a Jordan form associated with A in which all the blocks Jk (λk ) have distinct λk . 333 Exercise 12.5.7 (i) If A is an n × n matrix consisting of a single Jordan block with associated eigenvalue λ and ˙ x = Ax then writing y(t) = e−λt x(t) we have ˙ ˙ ˙ y(t) = −λe−λt x(t) + e−λt x(t) = −λe−λt x(t) + e−λt Ax(t) = B y(t) where B is a single Jordan block with associated eigenvalue 0. Thus y (n) = 0 and y (t) = n−1 j j =0 cj t for some cj ∈ R. Conversely, if y (t) = n−1 j j =0 cj t , for some cj ∈ R, then ˙ ˙ y(t) = B y(t). Thus n−1 x(t) = eλt cj tj j =0 where the cj ∈ R are arbitrary. (ii) Suppose that A is associated with a Jordan form B with blocks the nk × nk Jordan block Jk (λk ) with all the lambdak distinct [1 ≤ k ≤ r]. Writing A = M BM −1 with M non-singular we see that if y is a ˙ ˙ solution of y = B y if and only x = M y is a solution of x = Ax. Thus the only possible solutions of ⋆ are r eλk t Pk (t). x(t) = k=1 with the Pk general polynomials of degree nk − 1. To show that every x of this form is a solution observe that (by linearity) it suffices to show that eλk t Pk (t) is a solution with the Pk a general polynomials of degree nk − 1. By considering e−λk t x(t) we see that it suffices to show that if the characteristic equation of A has 0 as an p times repeated root the every polynomial of degree r − 1 satisfies ⋆. But if A has 0 as an p times repeated root then a0 = a1 = . . . = ar−1 = 0 so the previous sentence is automatic. 334 Exercise 12.5.8 (i) We have r r + k−1 k =k r! r! r! +(r+1−k )k = (r+1) = k !(r + 1 − k )! k !(r + 1 − k )! k !(r + 1 − k )! (ii) If k = 0 our system becomes ur (0) = ur−1 (0) with the general solution ur (0) = b0 . If k = 1 we have ur (0) = b0 as before and ur (1) − ur−1 (1) = b0 If we set vr (1) = ur (1) − b0 r we obtain vr (1) − vr−1 (1) = 0 so as before vr (1) = b1 for some freely chosen b1 . Thus ur (1) = b0 r + b1 . Suppose that the solution for a given k ≥ 0 is k ur (k ) = bk −j j =0 r . j Then k ur (k + 1) − ur−1 (k + 1) = bj j =0 r , j and setting k vr (k + 1) = ur (k + 1) − bk −j j =0 r , j+1 we obtain vr (k + 1) − vr−1 (k + 1) = 0 so, as before, vr (k + 1) = bk+1 for some freely chosen bk+1 . Thus k+1 ur (k + 1) = bk+1−j j =0 r . j Thus, by induction, k ur (k ) = bk −j j =0 r . j r+1 k 335 (ii) If we set ur (k ) = λ−r vr (k ) then the u(k ) satisfy the equations in (i). Thus k r r vr (k ) = λ bk −j . j j =0 (iii) The equation n−1 ur + aj uj −n+r = 0 j =0 may be rewritten as ut+1 + Aut = 0 where ut is the column vector (ut , ut+1 , . . . , ut+n−1 ) and 0 1 0 ... 0 0 0 0 1 ... 0 0 . . . . . . . A= 0 0 0 ... 1 0 0 0 0 ... 0 1 −a0 −a1 −a2 . . . −an−2 −an−1 By Exercise 12.5.6 we know that we can associate A with a Jordan form B in which in which the q blocks Jk (λk ) are of size sk × sk and have distinct λk . Let B = M −1 AM with M invertible. By (ii) the general solution of vt+1 + B vt = 0 is q sk λt ek k vt k=1 so bk,sk −j j =0 q sk λt ek k ut = k=1 ck,sk −j j =0 r j r . j We have shown that ut must be of the form just given. Direct computation shows that any ut of the form just given is a solution. 336 Exercise 12.5.9 All roots characteristic equation the same. λ10 λ10 λ00 0 λ 0 , 0 λ 0 , 0 λ 1 . 00λ 00λ 00λ Characteristic equation has two distinct λ λ00 0 λ 0 , 0 0 00µ with λ = µ. Characteristic equation has λ λ00 0 µ 0 , 0 0 00ν with λ, µ, ν distinct. roots 10 λ 0 0µ three distinct roots 10 λ 0 0µ 337 Exercise 12.5.10 We have A4 − A2 = 0 so the minimal polynomial m must divide t4 − t2 = t2 (t − 1)(t + 1) Since (A − I )A = A2 − A = 0 the minimal polynomial cannot be 1, t, t − 1 or t(t − 1) (i) m(t) = t2 . The characteristic polynomial must be t5 . The Jordan blocks must be of size 2 × 2 or less. There are three possibilities J00 J0 J0 0 J 0 with J = 0, 00 00 000 (ii) m(t) = t + 1. The Jordan normal form is −I giving one possibility. (iii) m(t) = t2 − 1. The Jordan normal form is diagonal with r entries 1, 5 − r entries −1 [1 ≤ r ≤ 4] and the characteristic equation is (t − 1)r (t + 1)s There are four possibilities. (iii) m(t) = t(t + 1). The Jordan normal form is diagonal with r entries −1, 5 − r entries −0 [1 ≤ r ≤ 4] and the characteristic equation is (t + 1)r ts There are four possibilities. (iv) m(t) = t2 (t − e) with e = ±1. The Jordan normal form either has one block J and then r diagonal entries e and 3 − r diagonal entries 0 [1 ≤ r ≤ 3] giving a characteristic polynomial t5−r (t − e)r or two blocks J and a diagonal entry e giving a characteristic polynomial t4 (t − e) There are eight possibilities. (v) m(t) = t(t − 1)(t + 1). (iii) m(t) = t2 − 1. The Jordan normal form is diagonal with r entries 1, s entries −1, u entries 0 [r, s, u ≥ 1, r + s + u = 5] and the characteristic equation is (t − 1)r (t + 1)s tu There are six possibilities. (vi) m(t) = t2 (t − 1)(t + 1). The Jordan normal form has one block J and r diagonal entries 1, s diagonal entries −1 [r, s ≥ 1, r + s ≤ 3] and 3 − r − s diagonal entries 0. The characteristic equation is (t − 1)r (t + 1)s t5−r−s . There are three possibilities. There are 29 possible Jordan forms. 338 Exercise 12.5.11 t−1 0 0 t−2 0 = (t − 1)2 (t − 2)2 det(tI − M ) = (t − 1) det 1 0 0 t−2 Looking at the eigenvalue 1 10 0 1 0 −1 00 implies we have x 10 x y y 0 0 = 2 0 z z w w 02 x+z y −y + 2z 2w =x =y =z =w so M (x, y, z, w)T = (x, y, z, w)T ⇔ (x, y, z, w) = (t, 0, 0, 0) for general t. Thus e2 = (1, 0, 0, 0)T is an eigenvector eigenvalue 1. If (M − I )x = e then z=1 0=0 −y + z = 0 w=0 Thus writing e1 = (0, 1, 1, 0)T we have (M −I )e1 = e2 = 0 (M −I )2 e1 = 0. Looking at the eigenvalue 2 101 0 1 0 0 −1 2 000 implies we have x 0 x y 0 y = 2 z 0 z w 2 w x+z y −y + 2z 2w = 2x = 2y = 2z = 2w 339 so M (x, y, z, w)T = (x, y, z, w)T ⇔ (x, y, z, w) = (t, t, 0, s) for general t and s. Thus the eigenspace has dimension 2 and basis (1, 1, 0, 0)T , (0, 0, 0, 1)T . The characteristic polynomial is (t − 1)2 (t − 2)2 minimal polynomial is (t − 1)2 (t − 2). A Jordan form is given by 1100 0 1 0 0 J = 0 0 2 0 0002 An appropriate basis is (0, 1, 1, 0)T , (1, 0, 0, 0)T (1, 1, 0, 0)T , (0, 0, 0, 1)T and we can take M to be the matrix with these columns, that is to say 0110 1 0 1 0 M = 1 0 0 0 . 0001 340 Exercise 13.1.2 (i) The tables are +01 001 110 ×01 000 101 By inspection of the addition table x + x = 0 for all x. (ii) x = −x ⇒ (1 + 1)x = 0 ⇒ x = 0. 341 Exercise 13.1.3 (i) If c = 0 then 0 = c−1 (cd) = (c−1 c)d = d (ii) If a2 = b2 then (a + b)(a − b) = a2 − b2 = 0 so a + b = 0 and/or b − a = 0 i.e. a = −b and or a = b. (iii) By (ii) (and the fact that 1 + 1 = 0, the equation x2 = c has no solutions or two solutions or c = 0 and we have x = 0. Thus every element y is a square root (just look at y 2 ) but every square has two distinct square roots or is 0. Thus k is odd, there are (k − 1)/2 non zero squares and k + 1)/2 squares in all. (iv) If we work in Z2 , then 02 = 0 and 12 = 1. Every element is a square. (Note −1 = 1) 342 Exercise 13.1.4 Suppose x, y, z ∈ R λ, µ ∈ Q. ˙˙ ˙˙ ˙˙ (i) (x+y )+z = (x + y ) + z = x + (y + z ) = (x+y )+z = x+(y +z ). ˙ ˙ (ii) x+y = x + y = y + x = y +x. ˙ (iii) x+0 = x + 0 = x. ˙˙ ˙˙˙ (iv) λ×(x+y ) = λ(x + y ) = λx + λy = λ×x+λ×y. ˙ ˙˙˙ (v) (λ + µ)×x = (λ + µ)x = λx + µx = λ×x+µ×y. ˙ ˙˙ (vi) (λµ)×x = (λµ)x = λ(µx) = λ×(µ×x). ˙ ˙ (vii) 1×x = 1x = x and 0×x = 0x = 0. 343 Exercise 13.1.6 (i) If M AM −1 = D with D = λ0 0µ then t2 − a = det(tI − A) = det(tI − D) = (t − λ)(t − µ) so t2 = a has a solution. Suppose c2 = a = 0. The eigenvalues are ±c which are distinct so by our standard arguments A is diagonalisable. (ii) Our standard argument shows that A is not diagonalisable. (iii) We know that 2 has no square root in Q so by (i), no M exists. (iv) Since some a do not have square roots the corresponding a are not diagonalisable. (v) If A = M −1 DM with D diagonal then I = A2 = M −1 D2 M so D = I . Thus (since we work in Z2 , D = I so A = I which is absurd. 2 If we set e1 = (1, 1)T and e2 = (1, 0)T we have a basis and Ae1 = e1 , Ae1 = e1 + e2 , Thus setting M= we have M AM 1 = 11 10 10 . 11 344 Exercise 13.1.7 The tables are + (0, 0) (1, 0) (0, 1) (1, 1) (0, 0) (0, 0) (1, 0) (0, 1) (1, 1) (1, 0) (1, 0) (0, 0) (1, 1) (0, 1) (0, 1) (0, 1) (1, 1) (0, 0) (1, 0) (1, 1) (1, 1) (0, 1) (1, 0) (0, 0) × (0, 0) (1, 0) (0, 1) (1, 1) (0, 0) (0, 0) (0, 0) (0, 0) (1, 0) (1, 0) (0, 0) (1, 0) (0, 1) (1, 1) (0, 1) (0, 0) (1, 1) (1, 1) (1, 0) (1, 1) (0, 0) (0, 1) (1, 0) (0, 1) We check the axioms for a field. (i) (a, b) + (c, d) = (a + c, b + d) = (c, d) + (a, b). (ii) We have ((a, b) + (c, d)) + (e, f ) = (a + c, b + d) + (e, f ) = (a + c) + e, (b + d) + f = a + (c + e), b + (d + f ) . (iii) (a, b) + (0, 0) = (a, b) (iv) (a, b) + (a, b) = (0, 0) (v) We have (a, b) × (c, d) = (ac + bd, ad + bc + bd) = (ca + db, da + cb + db) = (c, d) × (a, b). (vi) We have (a, b) × (c, d) × (e, f ) = (ac + bd, ad + bc + bd) × (e, f ) = (ac + bd)e + (ad + bc + bd)f, (ac + bd)f + (ad + bc + bd)e + (ad + bc + bd)f = a(ce + df ) + b(de + cf + df ), a(de + cf + df ) + b(ce + df ) + b(de + cf + df ) = (a, b) × (c, d) × (e, f ) (vii) (a, b) × (1, 0) = (a, b). (viii) By inspection of the multiplication table we see that if (a, b) = (0, 0), we can find (c, d) with a × (a, b) × (c, d) = (1, 0). 345 (ix) We have (a, b) × (c, d) + (e, f ) = (a, b) × (c + e, d + f ) = a(c + e) + b(c + e), a(d + f ) + b(c + e) + b(d + f ) = (ac + bd, ad + bd + bc) + (ae + bf, af + bf + be) = (a, b) × (c, d) + (a, b) × (e, f ) (x) By inspection of the multiplication table, all the elements are squares. 346 Exercise 13.1.8 (i) Suppose 2 = 0. If A = B + C with B T = B and C T = −C then 2B = B + C + B − C = (B + C ) + (B + C )T = A + AT so B = 2−1 (A + AT ) and C = A − B = 2−1 (A − AT ). Conversely if B = 2−1 (A + AT ) and C = A − B = 2−1 (A − AT ) then B T = 2−1 (AT + AT T ) = B C T = 2−1 (AT − AT T ) = −C B + C = 2−1 (A + AT ) + 2−1 (A − AT ) = A (ii) Suppose 2 = 0. Then a + a = (1 + 1)a = 2a = 0a = 0 so a = −a and (aij ) = (−aij ) so A = −A. Thus A = AT ⇔ A = −AT . We work over Z2 . If A = B + C with B T = B , C T = C then AT = A. Thus 01 00 cannot be written as the sum of a symmetric and an antisymmetric matrix. On the other hand the zero matrix 0 and the identity matrix I are symmetric so antisymmetric and 0 = 0 + 0 = I + I. 347 Exercise 13.1.9 (i) Observe that 02 + 0 = 0 + 0 and 12 + 1 = 1 + 1 = 0 so x2 = x = 0 for all x ∈ Z2 . (ii) If G is a finite field there are only a finite number of distinct functions f : G → G but there are infinitely many polynomials. Thus there must exist distinct polynomials P and Q which give rise to the same function. Consider P − Q. 348 Exercise 13.2.1 if. If U is a vector space over G then W ⊆ U is a subspace if and only (i) 0 ∈ W . (ii) x, y ∈ W → x = y ∈ W . (iii) λ ∈ G, x ∈ W ⇒ λx ∈ W . If G = Z2 then, if λ ∈ G, x ∈ W either λ = 0 and λx = 0 so, if condition (ii) is true λx ∈ W , or λ = 1 and λx = x ∈ W . Thus we can drop condition (iii) in this case. 349 Exercise 13.2.2 If U is infinite then all three statements are false. If U is finite then (since U spans itself) it has a finite spanning set and so a basis e1 , e2 , . . . , em . The elements of U are the vectors m xj ej j =1 with xj ∈ Z2 , each choice of the xj giving a distinct vector so U has 2m elements. If U is isomorphic to Zq they must have the same number of elements 2 so q = m. The map θ : Zm → U given by 2 m θ(x1 , x2 , . . . , xm ) = xj ej . j =1 Thus the three given statements are equivalent. 350 Exercise 13.2.3 We have n α(λe + µf ) = n n pj (λej + µfj ) = λ j =1 pj ej + µ j =1 pj fj = λα(e)+ µα(f ) j =1 so α is linear. Let e(j ) be the vector with 1 in the j th place and 0 elsewhere. If α is linear, then, setting pj = αe(j ) we have n n ej e(j ) α(e) = α j =1 = n ej αe(j ) = j =1 pj ej . j =1 351 Exercise 13.2.4 Codeword (0, 0, 0, 1, 1, 1, 1). Error in fourth place gives (0, 0, 0, 0, 1, 1, 1). z1 = 0, z2 = 0, z3 = 1 tells us error in 4th place. Recover (0, 0, 0, 1, 1, 1, 1). 352 Exercise 13.2.5 The tape will be accepted if each line contains an even number of errors. Since the probability of errors is small (and the number of bits on each line is small) the probability of one error is much greater than the probability of an odd number of errors greater than 1. Thus Pr(odd number of errors in one line) ≈ Pr(exactly one error in one line) = 8 × 10−4 × (1 − 10−4 )7 ≈ 8 × 10−4 . Since the the probability λ of an odd number of errors in one line is very small but there are a large number N of lines, we may use the Poisson approximation to get 1 − Pr(odd number of errors in some line) ≈ e−λN ≈ e−8 ≈ 0.00034 and conclude that the probability of acceptance is less than .04%. If we use the Hamming scheme, then, instead of having 7 freely chosen bits (plus a check bit) on each line, we only have 4 freely chosen bits (plus three check bits plus an unused bit) per line so we need approximately 1 × 7 × 104 = 1.75 × 104 4 lines. If a line contains at most one error, it will be correctly decoded. A line will fail to be correctly decoded if it contains exactly two errors and it may fail to be correctly decoded if it contains more than two errors. Since the probability of errors is small (and the number of bits on each line is small), the probability of two errors is much greater than the probability of more than two. Thus Pr(decoding failure for one line) ≈ Pr(exactly two errors in one line) = 7 × (10−4 )2 × (1 − 10−4 )5 ≈ 21 × 10−8 . 2 Since the the probability of a decoding error in one line is very small but there are a large number of lines, we may use the Poisson approximation to get Pr(decoding error for some line) = 1 − Pr(no decoding error in any line) ≈ 1 − e−21×10 −8 ×17500 ≈ 1 − e−.003675 ≈ .9963 and conclude that the probability of a correct decode is greater than 99.6%. 353 Exercise 13.2.6 (i) Observe that c1 c2 1 0 1 0 1 0 1 c3 0 0 1 1 0 0 1 1 c4 = 0 0 0 0 1 1 1 1 c5 0 c 6 c7 can be rewritten as c1 + c3 + c5 + c7 ≡ 0 c2 + c3 + c6 + c7 ≡ 0 c4 + c5 + c6 + c7 ≡ 0. (ii) Observe that AeT is the j row of A and that the j th row of A is j the binary digits of j written in reverse order. By linearity A(c + ej )T = AcT + AeT = AeT = (a1 , a2 , a3 )T so the j j statement follows. 354 Exercise 13.2.7 AeT = AeT for i = j since otherwise an error in the j th place would i j have the same effect as an error in the ith place. Since x+y =0⇔x=y we have A(eT + eT ) = 0T i j Thus by linearity A(c + ei )T + ej )T ) = Ac + AcT + AeT j = A(eT + eT ) = 0 i j and c + ei )T + ej is not a codeword. (ii) Aej ∈ Z3 \ {0}. Since the Aej are distinct for distinct j and {j : 1 ≤ j ≤ 7} has the same number of elements as Z3 \ {0}T we have bijection and so since A(eT + eT ) = 0 i j there must be a k with A(eT + AeT ) = A(eT ). i j k If k = j we would have AeT = 0T which is impossible. Thus j = k and i similarly j = i. If we have an error in the i and j th place the Hamming system will assume an error in the k th place. 355 Exercise 13.2.9 Observe that if A = (aij ) c ∈ C ⇔ AcT = 0T n ⇔ where αj ∈ (Zn )′ 2 j =1 aij cj = 0 for 1 ≤ j ≤ r ⇔ αj c = 0 for 1 ≤ j ≤ r is given by n αj c = aij cj . j =1 356 Exercise 13.2.10 The null space of a linear map is a subspace. 357 Exercise 13.2.12 Let α be the linear map associated with A for the standard bases. Since A has rank r so does α and dim C = dim α−1 (0) = n − rank(α) = n − r so C has 2n−r elements. 358 Exercise 13.2.13 (i) Suppose C has basis g(j ) [1 ≤ r ≤ n]. Consider an r × n matrix U with j th row g(j ). We can use elementary elementary row operations so that (after reordering coordinates if necessary) we get V = I B . Since elementary row operations leave the space spanned by the rows of a matrix unchanged, C has basis the rows e(j ) of V . This gives the required result. (ii) The rank nullity theorem tells us that the Hamming code has dimension 4. If we do not interchange columns then considering the effect of choosing one of the c3 , c5 , c6 , c7 to be 1 and the rest 0 we obtain 4 independent vectors. (1, 1, 1, 0, 0, 0, 0), (1, 0, 0, 1, 1, 0, 0) (0, 1, 0, 1, 0, 1, 0), (1, 1, 0, 1, 0, 0, 1) which therefore form a basis. (iii) α is linear and surjective so (since the spaces have the same dimension) an isomorphism. λx + µy, β (λx + µy) = α(λx + µy) = λαx + µαy = λ(x, β x) + µ(y, β y) = (λx + µy, λβ x + µβ y) so β (λx + µy) = λβ x + µβ y and β is linear. (iv) Neither is true. Take n = 2r b(j ) = 0. 359 Exercise 13.2.14 The probability of exactly 2 errors in a line (ie codeword), in which case the correction is erroneous, is 7 2 1 10 2 9 10 5 ≈ 0.124 The probability of no such problem in 17 500 lines is less than (9/10)17500 which is negligible. 360 Exercise 13.2.15 If p = 1/2 then, whatever the message sent, the received sequence is a series of independent random variables each taking the value 0 with probability 1/2 and the value 1 with probability 1/2. If p > 1/2, then replacing each received 0 by 1 and each received 1 by 0 gives a system for the probability of an error in one bit is 1 − p. 361 Exercise 13.2.16 By definition 10101)1 101 , A3 = 0 1 1 0 0 1 1 011 0001111 A1 = (1), A2 = and 1 0 A4 = 0 0 0 1 0 0 1 1 0 0 0 0 1 0 1 0 1 0 0 1 1 0 1 1 1 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 0 1 0 0 1 1 1 0 1 1 0 1 1 1 1 1 1 1 (ii) The 2r th column has a 1 in the rth place with all other entries zero, so these columns are linearly independent and A has rank n. +1 (iii) Let us write j = n=1 aij 2i−1 with aij ∈ {0, 1}. Then, if 0 ≤ i j < 2n , aij = ai(j +2n ) for 1 ≤ i ≤ n − 1 and an,j = 0, an,(j +2n ) = 1. We have ai2n−1 = δin . 362 Exercise 13.2.17 Let ek be a row vector of length 2n − 1 with 1 in the k th place an n− zero elsewhere. Write k = i=11 bi 2i−1 with bi ∈ {0, 1}. Then, if c is a code word, A(c + ek )T = AeT = b, k so we can recover from a single mistake by the rules:(1) If x received and AxT = 0T , assume message correct. (2) If x received and AxT = bT , assume mistake in place. n i−1 th i=1 bi 2 Since A has full rank (see previous question), the rank-nullity theorem tells us that C has 2n−r elements. 363 Exercise 13.2.18 Since the error rate is very low we can use the Poisson approximation. (But exact calculation is easy using a calculator.) We have a rate (5 × 107 ) × 10−7 = 10 so the probability that no error occurs is roughly e−5 which is very small. By the Poisson approximation (or directly by bounding terms) the probability of more than one error in a line is roughly 1 p = λ2 e−λ 2! −7 with λ = 63 × 10 . Thus p ≈ 2 × 10−10 There are roughly 105 lines so the probability of an error in the decoded message is (using the Poisson approximation) less than about exp(−2 × 10−5 ) which is pretty small. 364 Exercise 13.3.3 C ([a, b]) is a subset of the vector space R[a,b] of functions f : [a, b] → R. We observe that 0 ∈ C ([a, b]) and that f, g ∈ C ([a, b]) ⇒ λf + µg ∈ C ([a, b]) so C ([a, b]) is a subspace of R[a,b] and so a vector space. We have b f, f = a f (x)2 dx ≥ 0 and since the integral of a positive continuous function is zero if and only if the function is zero b f, f = 0 ⇒ Also a f (x)2 dx = 0 ⇒ f × f = 0 ⇒ f = 0. b b g (x)f (x) dx = g , f , f (x)g (x) dx = f, g = a a b f (x)(g (x) + h(x) dx f, g + h = a b = f (x)g (x) + f (x)h(x) dx a = f, g + f, g and b b f (x)g (x) dx = λ f , g . λf (x)g (x) dx = λ λf, g = a a 365 Exercise 13.3.4 (i) If x = 0 or y = 0, we have equality. If λ is a real number then we have 0 ≤ λx + y 2 = λx + y , λx + y = λ2 x = 2 2 + 2λ x, y + y x, y x λx + 2 2 +y If we set λ=− − x, y x 2 x·y x 2 . x, y , x2 we obtain 2 0 ≤ (λx + y) · (λx + y) = y − . Thus y ⋆ 2 2 x, y x − ≥0 with equality only if 0 = λx + y so only if λx + y = 0. Rearranging the terms in ⋆ we obtain x, y and so with equality only if 2 ≤x 2 y 2 | x, y | ≤ x y λx + y = 0. We observe that, if λ′ , µ′ ∈ R are not both zero and λ′ x = µ′ y, then | x, y | = x y . Thus | x, y | ≤ x y with equality if and only if x and y are linearly dependent. (ii) By definition x ≥ 0. Further x ≥ 0 ⇔ x, x = 0 ⇔ x = 0. Since λx, λx = λ2 x, x , we have λx = |λ| x . 366 Finally 2 =x 2 + 2 x, y + y ≤x x+y 2 2 +2 x y + y 2 = ( x + y )2 , so x + y ≤ x + y . (iii) If we use the inner product of Exercise 13.3.3, the Cauchy– Schwarz inequality takes the form 2 b f (t)g (t) dt a b ≤ b f (t)2 dt a g (t)2 dt. a 367 Exercise 13.3.5 (i) We have p0 = 1, so the result is true for n = 0. If pr is polynomial of degree r for r ≤ n − 1, then n−1 j =0 qn , p j pj 2 is a polynomial of degree at most n − 1, so n−1 p n = qn − j =0 qn , p j pj pj 2 is a monic polynomial of degree n. p1 (x) = x − 1 qn , p j pj (x) pj 2 2 p1 (x) = x − q1 , p 0 p0 (x) = x. p0 2 j =0 1 = x2 − 1 t2 dt 12 dt −1 −1 = x2 − 1 . 3 (ii) 0 is a polynomial of degree less than n. If P and Q are polynomials of degree at most n so is λP = µQ. Writing qr (x) = xr , we have the trivial equality n n r ar x = r =0 ar q r (x) r =0 so the qj with 0 ≤ j ≤ n span Pn . Since a non-zero polynomial of degree at most n can have at most n zeros n r =0 ar qr = 0 ⇒ ar = 0 [0 ≤ r ≤ n] so the qr form a basis and Pn has dimension n + 1. The pj pj −1 form a collection of n + 1 orthonormal vectors so form an orthonormal basis. (iii) If P is monic of degree n + 1, then P = pn+1 + Q 368 with Q a polynomial of degree at most n Now P ⊥ Pn ⇒ (P − pn+1 ) ⊥ Pn ⇒ Q ⊥ Pn ⇒ Q ⊥ Q ⇒ Q, Q = 0 ⇒ Q = 0 ⇒ P = pn+1 . By definition pn+1 is monic and pn+1 ⊥ span{p0 , p1 , . . . , pn } and so, since Pn span{p0 , p1 , . . . , pn }, we have pn+1 ⊥ Pn . 369 Exercise 13.3.11 (i) Observe that n j =1 Similarly 2n j =n+1 n |λj | ≥ 1 λj = 1 dt = 2. −1 j =1 |λj | ≥ 2. (ii) We have 2n n µj P (tj ) = (ǫ −1 + 1) j =1 j =1 2n λj P (tj ) − ǫ −1 1 for all polynomials P of degree n or less. Let f be the simplest piecewise linear functional with if 1 ≤ j ≤ n and λj ≥ 0 ǫ −ǫ if 1 ≤ j ≤ n and λ < 0 j f (tj = ǫ if n + 1 ≤ j ≤ 2n and λj < 0 −ǫ if n + 1 ≤ j ≤ 2n and λj ≥ 0 with f (−1) = 0 if −1 ∈ {tj : 1 ≤ j ≤ 2n} / if 1 ∈ {tj : 1 ≤ j ≤ 2n}. / f (1) = 0 Then |f (t)| ≤ ǫ for all t ∈ [−1, 1], but 2n 2n µj P (tj ) = j =1 j =1 P (t) dt λj P (tj ) = j =n+1 |µj | ≥ 2(1 + ǫ−1 ) + 2ǫ−1 . −1 370 Exercise 13.3.12 (i) Observe that b f, f = a f (x)2 r(x) dx ≥ 0, 2 and, since x → f (x) r(x) is continuous and positive b f, f = 0 ⇒ f (x)2 r(x) dx = 0 a ⇒ f (x)2 r(x) = 0 ∀x ⇒ f (x) = 0 ∀x Automatically b b g (x)f (x)r(x) dx = g , f f (x)g (x)r(x) dx = f, g = a a and b (λ1 f1 (x) + λ2 f2 (x))g (x)r(x) dx λ1 f1 + λ2 f2 , g = a b λ1 f1 (x)g (x) + λ2 f2 g (x)r(x) dx = a = λ1 f1 , g + λ2 f2 , g . (ii) This is just the Cauchy–Schwartz inequality f , f g, g ≥ f , g 2. (iii) We define Pn inductively. Set P0 = 1. Suppose that Pj is a monic polynomial of degree j with the Pj orthogonal with respect to our inner product [0 ≤ j ≤ n − 1]. Then, if Qn (x) = xn , n−1 Pn = Qn − j =0 Qn , Pj Pj , Pj is a monic polynomial of degree n orthogonal to all the Pj with j ≤ n − 1. Since the subspace Pn−1 of polynomials of degree at most n − 1 has dimension n and the Pj are non-zero and mutually orthogonal the Pj with j ≤ n − 1 form a basis for Pn−1 and Pn , Q = 0 for all polynomials Q of degree n − 1 or less. (iii) Let x1 , x2 , . . . xk be the real roots of Pn which have odd order and lie in (a, b). Setting Q(x) = k=1 (x − xk ), we see that Pn Q is j 371 single signed so b Pn (x)Q(x)r(x) dx = 0 Pn , Q = a and k ≥ n. Thus k = n and Pn has n distinct real roots all lying in (a, b). If we set ev (x) = j =v,1≤j ≤n (x b λv = − xj )(xv − xj )−1 and ev (x)r(x) dx a then (since f − n=1 f( xv )ev is a polynomial of degree at most n − 1 v vanishing at n points) we have f = n=1 f (xv )ev and v n b f (x)r(x) dx = a λv f( xv ) v =1 for all f polynomials of degree n − 1 or less. If f is polynomials of degree 2n − 1 or less, then f = gPn + h with g and h polynomials of degree n − 1 or less, so b b b a a a h(x)r(x) dx g (x)Pn r(x) dx + f (x)r(x) dx = n =0+ λ v g( x v ) v =1 n = n λ v g( x v ) + v =1 n = λv 0 v =1 n λ v g( x v ) + v =1 n = v =1 λv f (xv ). v =1 λv Pn (xv )h(xv ) 372 Exercise 13.3.14 We need to show that Pn , qPn−1 > 0. Now qPn−1 and Pn are both monic polynomials of degree n, so qPn−1 − Pn has degree at most n − 1. Thus and Pn , qPn−1 − Pn = 0 Pn , qPn−1 = Pn , Pn > 0. 373 Exercise 13.3.16 Observe that 2−8n 1 0≤ 2 n+1 fn (x) dx = (2 1 as n → ∞ so fn → 0. fn (x)2 dx − 1) −2−8n n+1 −8n+1 2n −5n+2 ≤2 2 2 =2 →0 However, if u is an integer with |u| < 2m − 1, then f (u2−m ) = 2n when n ≥ m and so fn (u2−m ) → ∞ as n → ∞. 374 Exercise 13.4.1 If dim U = 0, there is nothing to prove. Otherwise, we construct the ej inductively. Since dim U > 0 we can find a a = 0. Set e1 = a −1 a. If e1 , e2 , . . . , er have been found as orthonormal vectors either r = n and we have an orthonormal basis or we can find b ∈ span{e1 , e2 , . . . , er } / The Gram–Schmidt method now produces an er+1 with e1 , e2 , . . . , er+1 orthonormal. The process terminates in an orthonormal basis. Since the ej are linearly independent and span the map n θ = (x1 , x2 , . . . , xn )T xj e j j =1 is a well defined bijection. Since n θλ n n xj ej + µ j =1 yj e j =θ (λxj + µyj )ej j =1 j =1 = (λx1 + µy1 , λx2 + µy2 , . . . , λxn + µyn )T = λ(x1 , x2 , . . . , xn )T + µ(y1 , y2 , . . . , yn )T n n = λθ λ xj ej + µθ yj e j j =1 j =1 θ is linear. Finally, n n θ j =1 xj ej ) ·θ yj e j j =1 = xj yj = j =1 n n n xj e j θ j =1 yj e j ,θ j =1 . 375 Exercise 13.4.3 Choose an orthonormal basis e1 , e2 , . . . , e r for V . Then by Bessel’s inequality r a− λj ej j =1 has a unique minimum when λj = a, ej . If x ∈ V a − x, v = 0 ∀v ∈ V ⇔ a − x, ej = 0 ∀1 ≤ j ≤ r ⇔ x, ej = a, ej ∀1 ≤ j ≤ r r ⇔x= a, ej ej j =1 so we are done. Exercise 13.4.5⋆ 376 Exercise 13.4.6 Observe that n u, a = α(ej )ej , u j =1 n = α(ej ) ej , a j =1 n ej , a ej =α j =1 = αu. 377 Exercise 13.4.7 We have a map α : U → R so we only need check linearity. But, if u, v ∈ U and λ, µ ∈ R we have α(λu + µv) = λu + µv, a = λ u, a + µ v, a = λαu + µαv, so we are done. 378 Exercise 13.4.10 Observe that u, Φ(λα + µβ )v = (λα + µβ )u, v = λα(u) + µβ (u), v = λ αu, v + µ β u, v = λ u, α∗ v + µ u, β ∗ v = u, (λα∗ + µβ ∗ )v = u, (λΦα + µΦβ )v for all u ∈ U , so for all v ∈ U so Φ(λα + µβ )v = (λΦα + µΦβ )v Φ(λα + µβ ) = λΦα + µΦβ for all α, β ∈ L(U, U ) and λ, µ ∈ F, so Φ is linear. 379 Exercise 13.4.12 D is a subset of the inner product space C ([0, 1]). 0 ∈ D and λ, µ ∈ R, f, g ∈ D ⇒ λf + µg ∈ D. Thus D is a subspace of the inner product space C ([0, 1]) and so an inner product space. Further, α(λf + µg ) = p(λf + µg )′ ′ = λpf ′ + µgf ′ ′ = λαf + µαg. so α is linear. By integration by parts, 1 (f ′ p)′ (t)g (t) dt αf, g = 0 = f ′ (t)p(t)g (t) 1 0 1 − f ′ (t)p(t)g ′ (t) dt 0 1 =− f ′ (t)p(t)g ′ (t) dt 0 = αg, f = f , αg , ∗ so α = α. 380 Exercise 13.5.2 We have z, λw P = λw , z ∗ P = λ∗ w , z ∗ = λ w, z P ∗ P = λ z, w P . Let , M be defined as in the question. (i) Since z, z positive, so is z, z M = z, z ∗ . P (ii) z, z M (iii) λz, w = 0 ⇒ z, z M = λz, w P ∗ P = 0 ⇒ z = 0. = λ∗ z, w ∗ P = λ z, w ∗ P P is real and = λ z, w M. (iv) We have z + u, w M = z + u, w = z, w P = z, w = z, w (v) w, z M = w, z ∗ P ∗ P M = z, w ∗∗ P ∗ P ∗ + u, w + u, w ∗ P + u, w = z, w P M ∗ M. Essentially same arguments show that if , inner product then z, w P = z, w ∗ M defines a physicist’s inner product. M is a mathematician’s 381 Exercise 13.5.3 C ([a, b]) is a subset of the vector space C[a,b] of functions f : [a, b] → C. We observe that 0 ∈ C ([a, b]) and that f, g ∈ C ([a, b]) ⇒ λf + µg ∈ C ([a, b]) so C ([a, b]) is a subspace of R[a,b] and so a vector space. We have b f, f = a |f (x)|2 dx ≥ 0 and since the integral of a positive continuous function is zero if and only if the function is zero b f, f = 0 ⇒ |f (x)|2 dx = 0 ⇒ f × f = 0 ⇒ f = 0. a Also b b (g (x)f (x)∗ )∗ dx ∗ f (x)g (x) dx = f, g = a a ∗ b g (x)f (x)∗ dx = = g, f , a and b f (x)(g (x) + h(x)∗ dx f, g + h = a b f (x)g (x)∗ + f (x)h(x)∗ dx = f , g + f , g , = a whilst b b f (x)g (x)∗ dx = λ f , g . ∗ λf (x)g (x) dx = λ λf, g = a a 382 Exercise 13.5.4 (i) We have n f , ej = ar e r , e j r =1 n = n ar e r , e j = r =1 Note that ej , f = f , e j ar δrj = aj r =1 ∗ = a∗ . j (ii) We have 1 1 er , es = 0 exp(2πirt) exp(−2πist) dt = 0 exp 2πi(r−s)t dt = δrs . 383 Exercise 13.5.5 (i) Observe that k v, er = x, er − x, er ej , er = x, er − x, er = 0 j =1 for all 1 ≤ r ≤ k . (ii) If v = 0, then k x= j =1 x, ej ej ∈ span{e1 , e2 , . . . , ek }. If v = 0, then v = 0 and k x=v+ x, ej ej j =1 k = v ek+1 + j =1 x, ej ej ∈ span{e1 , e2 , . . . , ek+1 }. (iii) If e1 , e2 , . . . , ek do not form a basis for U , then we can find x ∈ U \ span{e1 , e2 , . . . , ek }. Defining v as in part (i), we see that v ∈ U and so the vector ek+1 defined in (ii) lies in U . Thus we have found orthonormal vectors e1 , e2 , . . . , ek+1 in U . If they form a basis for U we stop. If not, we repeat the process. Since no set of n + 1 vectors in U can be orthonormal (because no set of n + 1 vectors in U can be linearly independent), the process must terminate with an orthonormal basis for U of the required form. 384 Exercise 13.5.7 Let V have an orthonormal basis e1 , e2 , . . . , en−1 . By using the Gram–Schmidt process, we can find en such that e1 , e2 , . . . , en are orthonormal and so a basis for U . Setting b = en , we have the result. 385 Exercise 13.5.8 Uniqueness. If αu = u, a1 = u, a2 for all u ∈ U , then u, a1 − a2 = 0 for all u ∈ U and, choosing u = a1 − a2 , we conclude, in the usual way, that a1 − a2 = 0. Existence. If α = 0, then we set a = 0. If not, then α has rank 1 (since α(U ) = R) and, by the rank-nullity theorem, α has nullity n − 1. In other words, α−1 (0) = {u : αu = 0} has dimension n − 1. By Lemma 13.4.2, we can find a b = 0 such that that is to say, If we now set α−1 (0) = {x ∈ U : x, b = 0}, α(x) = 0 ⇔ x, b = 0 a= b −2 α(b)b, we have and αx = 0 ⇔ x, a = 0 αa = b −2 α(b)2 = a, a . Now suppose that u ∈ U . If we set αu x=u− a, αa then αx = 0 so x, a = 0, that is to say, αu αa 0= u− a, a = u, a − αu = u, a αa a, a and we are done. 386 Exercise 13.5.9 Since λu + µv, a = λ u, a + µ u, a θ(a ∈ U . The previous exercise tells us that θ is surjective. ′ θ(a) = θ(b) ⇒ b − a, b − a ⇒ a = b so θ is injective. Finally, θ(λa + µb)(u) = u, λa + µb = λ∗ u, a + µ∗ u, b = λ∗ θ(a)u + µ∗ θ(b)u = λ∗ θ(a) + µ∗ θ(b) u for all u ∈ U . Thus θ(λa + µb) = λ∗ θ(a) + µ∗ θ(b) as required. Exercise 13.5.10⋆ 387 Exercise 13.5.11 Observe that, if v ∈ U , the map θ(u) = αu, v lies in U ′ , so the Riesz representation theorem tells us that there is a unique α∗ v ∈ U such that αu, v = u, α∗ v for all v ∈ U . Since u, α∗ (λv + µw) = αu, (λv + µw) = λ∗ αu, v + µ∗ αu, w = λ∗ u, α∗ v + µ∗ u, α∗ w = u, λα∗ v + µα∗ w for all u ∈ U , we have α∗ (λv + µw) = λα∗ v + µα∗ w so α∗ ∈ L(U, U ). Again, u, (λα + µβ )∗ v = (λα + µβ )u, v = λ αu, v + µ β u, v = λ u, α∗ v + µ u, β ∗ v = u, λ∗ α∗ v + µ∗ β ∗ v = u, (λ∗ α∗ + µ∗ β ∗ )v for all u ∈ U , so (λα + µβ )∗ v = (λ∗ α∗ + µ∗ β ∗ )v for all v ∈ U , so (λα + µβ )∗ = λ∗ α∗ + µ∗ β ∗ and he map Ψ : L(U, U ) → L(U, U ) given by Ψα = α∗ satisfies Ψ(λα + µβ ) = λ∗ Ψα + µ∗ Ψβ. Finally, α∗∗ u, v = v, α∗ u = α∗ v, u ∗ = u, α∗ v = αu, v for all v ∈ U , so for all u ∈ U , so α∗∗ anti-isomorphism. α∗∗ u = u = α and Ψ is its own inverse, so bijective so an 388 Exercise 13.5.12 If α has matrix A = (aij ) and α∗ has matrix B = (bij ) with respect to the given basis, then n brj = = bij ei , er i=1 α ∗ ej , er = e r α ∗ ej ∗ n = akr ek , ej k=1 = a∗ , jr so B = A∗ . ∗ = α er , ej 389 Exercise 13.5.13 (i) Since V and V ⊥ are complementary, dim V + dim V ⊥ = dim U. For the same reason, dim V ⊥ + dim V ⊥⊥ = dim U , so dim V ⊥⊥ + dim V . Since x ∈ V ⇒ x, y = y, x = 0 ∀y ∈ V ⊥ we have V ⊆ V ⊥⊥ so V = V ⊥⊥ . (ii) The statement is equivalent to saying that, if u ∈ U , there are unique a ∈ V and b ∈ V ⊥ such that u = a + b. and this is the same as saying that V and V ⊥ are complementary. (iii) We have λ1 u1 + λ2 u2 = (λ1 π u1 + λ2 π u2 ) + (λ1 (ι − π )u1 + λ2 (ι − π )u2 ) and λ1 π u1 + λ2 π u2 ∈ V , λ1 (ι − π )u1 + λ2 (ι − π )u2 ∈ V ⊥ so, by uniqueness, π (λ1 u1 + λ2 u2 ) = λ1 π u1 + λ2 π u2 . Thus π and so ι − π are linear. Since π u ∈ V we have, by uniqueness π 2 U = π (π (u) = π u for all u ∈ U so π = π 2 . If x, y ∈ U , then π x, y = π x, π y + (ι − π )y = π x, π y + π x, (ι − π )y = π x, π y = π x, π y + (ι − π )x, π y = x, π y so π = π ∗ 390 Exercise 13.5.14 (i)⇒(ii) Let U = α(V ) and W = (ι − α)(V ). Since v = αv + (ι − α)v U + W = V . If u ∈ U then u = αv for some v ∈ V so αu = α2 v = αv = u = ιU u. If w ∈ W then u = (ι − α)v for some v ∈ V so αw = (α − α2 )v = 0. Finally, if u ∈ U , w ∈ W the results of the previous paragraph give u, w = αu, (ι − α)w = u, α(ι − α)w = u, 0 = 0. (ii)⇒(iii) Immediate. (iii)⇒(iv) Take an orthonormal basis for U and an orthonormal basis for W . Together they form an orthonormal basis for V with the required properties. (iv)⇒(i) Let A be the matrix of α with respect to the specified basis. Then A∗ = A and A2 = A, so the results for α follow. An orthogonal projection automatically obeys the condition α2 = α. Give C2 the standard inner product. Let 2 2 α(z, w) = ( 1 z + 3 w, 1 z + 3 w). 3 3 By inspection, α is linear and α2 = α. By observing that the matrix A of α with respect to the standard basis is real, but not symmetric, we see that A∗ = A so α∗ = α. Thus α is a projection but not an orthogonal projection. If α and β are orthogonal projections and αβ = βα, then (αβ )2 = (αβ )(αβ ) = α(βα)β = α(αβ )β = α2 β 2 = αβ and (αβ )∗ = (βα)∗ = α∗ β ∗ = αβ . 391 Exercise 13.5.15 We have ρW x 2 2 ρ2 = 4πW − 4πw + ι = ι. W = (2πW − ι)x, (2πW − ι)x = (2πW − ι)x, (2πW − ι)x = x, x = x 2 so ρW is an isometry. ρW ρW ⊥ = (2πW − ι)(2πW ⊥ − ι) = 4πW πW ⊥ − 2πW − 2πW ⊥ + ι = 0 − 2ι + ι = −ι. 392 Exercise 14.1.1 d1 (A, B ) = maxi,j |aij − bij | = A − B 1 . (i) |aij | ≥ 0 ∀i, j so A (ii) A 1 1 ≥ 0. = 0 ⇒ |aij | = 0 ∀i, j ⇒ aij = 0 ∀i, j ⇒ A = 0. (iii) λA = maxi,j |λaij | = maxi,j |λ||aij | = |λ| maxi,j |aij | = |λ A 1 . 1 (iv) |aij − bij | ≤ |aij | + |bij | ∀i, j so max |aij − bij | ≤ max |aij | + max |bij | i,j and A + B 1 i,j ≤A d2 (A, B ) = i,j 1 i,j + B 1. |aij − bij | = A − B 2 . (i) |aij | ≥ 0 ∀i, j so A 2 = i,j (ii) A 2 |aij | ≥ 0. = 0 ⇒ |aij | = 0 ∀i, j ⇒ aij = 0 ∀i, j ⇒ A = 0. (iii) We have λA 2 = i,j = |λ| |λaij | = i,j i,j |λ||aij | |aij | = |λ| A 2 (iv) We have A+B 2 = i,j = i,j d3 (A, B ) = n2 . 3 |aij + bij | ≤ |aij | + 2 i,j |aij − bij | 1/2 i,j i,j |aij | + |bij | |bij | = A 2 + B 2. = A − B 3. is just the standard norm for an inner product space of dimension 393 Exercise 14.1.3 Observe that, using the Cauchy–Schwarz inequality, 1/2 n αx = 2 n aij xj i=1 n j =1 n = i=1 j =1 |aij |2 x n 2 n ≤ i=1 j =1 1/2 n ≤ 1/2 n 2 |aij | j =1 |xj | 1/2 n i=1 j =1 2 |aij |2 x. 394 Exercise 14.1.7 Since U = (ι−α)−1 (0) and W = α−1 (0) are orthogonal complements, we can write any v ∈ V in the form v =u+w with u ∈ U and w ∈ W . Since u ⊥ w, v 2 =u 2 +w 2 and so π ≤ 1. πv = u ≤ v Since π = 0, we can find x with u = π x = 0 so u = 0, Since π (u = u, we have π = 1. We use row vectors. Let K > 0. If α(x, y ) = (x + Ky, 0), then α is linear and α2 (x, y ) = (x + Ky, 0) = α(x, y ), so α is a projection. Since (0, 1) = 1 and α(0, 1) = K , α ≥ K . 395 Exercise 14.1.9 If e1 , e2 , . . . , en is an orthonormal basis for U , then the map n θ xj e j = (x1 , x2 , . . . , xn )T j =1 is a linear inner product preserving (and so norm preserving) isomorphism θ : U → Fn where Fn is equipped with the standard dot product. 396 Exercise 14.1.10 Observe that Ix = x so I = 1. (i) I + (−I ) = 0 = 0, I = − I = 1. (ii) I + I ) = 2I = 2, I = 1. (iv) I I = I = 1. We now attack (iii). If J= 01 , 00 then J (x, y )T = (y, 0)T = |y | ≤ (x, y )T so J | ≤ 1. However (1, 0)T = (0, 1)T = 1 and J (0, 1)T = (1, 0)T so J = 1 but J J = 0 = 0. 397 Exercise 14.1.11 det(tI − A) = (t − 1)2 − µ2 = t2 − 2t + η so the eigenvalues are 1 ± (1 − η 2 )1/2 The eigenvectors corresponding to 1 + (1 − η 2 )1/2 are given by x + µy = (1 + (1 − η 2 )1/2 )x µx + y = (1 + (1 − η 2 )1/2 )y i.e x = y (as we could spot directly). The eigenvectors corresponding to 1 − (1 − η 2 )1/2 are given by x + µy = (1 − (1 − η 2 )1/2 )x µx + y = (1 − (1 − η 2 )1/2 )y i.e x = −y (as we could spot directly). Thus we have a basis of orthonormal eigenvectors e1 = 2−1/2 (1, 1)T eigenvalue λ1 = 1 + (1 − η 2 )1/2 and e2 = 2−1/2 (1, −1)T eigenvalue λ2 = 1 − (1 − η 2 )1/2 . We have T (x1 e1 + x2 e2 = λ1 x1 e1 − λ2 x2 e2 = (λ1 x1 )2 + (λ2 x2 ) ≤ λ1 (x2 + x2 )1/2 1 2 1/2 ≤ λ1 (x1 e1 + x2 e2 so A ≤ λ1 (in fact, since Ae1 = λ1 e1 we have A = λ1 ) and small changes in x produce small changes in y. Observe that taking, y = e2 , we have x = A−1 = λ−1 which is very 2 large. (ii) Observe that looking at A−1 we have a basis of orthonormal eigenvectors e1 = 2−1/2 (1, 1)T eigenvalue λ1 = 1 + (1 − η 2 )−1/2 and e2 = 2−1/2 (1, −1)T eigenvalue λ2 = 1 − (1 − η 2 )−1/2 . Thus arguing as in (i), A−1 = λ−1 . 2 Now det B = det A det A−1 = det AA−1 = det I = 1 and A−1 0 . B −1 = 0A By looking at vectors of the form (0, 0, x, y ) we see that B ≥ A−1 and by looking at vectors of the form (x, y, 0, 0) we see that B ≥ A−1 . Thus if η is very small, both B and B −1 are very large. 398 Exercise 14.1.12 (i) We have c(A) = A A−1 | ≥ AA−1 = I = 1. (ii) We have c(λA) = λA (λA)−1 = λA λ−1 A−1 | = |λ||λ−1 | A A−1 = c(A). 399 Exercise 14.1.13 Let us use an orthonormal basis e1 , e2 , . . . , en . Observe that n− 2 i,j |aij | ≤ n−2 max |ars | = max |ars |. r,s i,j r,s and that 1/2 n 2 |ars | ≤ |ajs | j =1 n = ajs ej j =1 If x = = α es ≤ α n j =1 es = α . xj ej then |xj | = | x, ej | ≤ x so n n αx = r =1 xr α(er ) ≤ r =1 |xr | α(er ) n ≤x ≤x n n α(er ) = x r =1 n ajr ej r =1 j =1 n n n r =1 j =1 |ajr | ej ≤ x r =1 j =1 |ajr |, so α≤ |aij |. i,j Finally, i,j |aij | ≤ i,j max |ars | = n2 max |ars |. r,s r,s Since n 1/2 n 2 αx ≤ i=1 j =1 |aij | for all x ≤ 1, we have n αy ≤ 1/2 n 2 i=1 j =1 |aij | y for all y and 1/2 α≤ a2 ij i,j . 400 Exercise 14.2.2 A triangular matrix A has characteristic polynomial n χA (t) = j =1 (t − aii ) so, if A is real, its characteristic equation has all its roots real. Thus, if α is triangularisable, its characteristic equation has all its roots real. Consider the matrix C0 B= 0I with I the n − 2 × n − 2 identity matrix and C= 0 −1 10 B has characteristic polynomial (t − 1)n−2 (t2 +1) so not all its roots are real. If β is the linear map corresponding to B with respect to some basis then β is not triangularisable. If dim V = 1 a 1 × 1 matrix is triangular so all endomorphisms are triangularisable. The fact that A = QR with R upper triangular and Q orthogonal does not imply that there is an invertible matrix P and an upper triangular matrix T with A = P AP −1 . 401 Exercise 14.2.4 Let A be the matrix of an endomorphism α with respect to some orthonormal basis. We can find α(n) ∈ L(U, U ) with α(n) − α → 0 as n → ∞ such that the characteristic equations of the α(n) have no repeated roots. Let A(n) be the matrix of α(n) with respect to the given basis. Then max |aij (n) − aij | ≤ α(n) − α → 0 i,j as n → ∞. 402 Exercise 14.3.2 We have det(tI − A) = (t − 1/2)(t − 1/4), so A has eigenvalues 1/2 and 1/4 By induction on n, An (0, 1)T = (2−n+1 K, 0)T for all n ≥ 1. Thus, if K = 2N L, we have AN (0, 1)T > L. 403 Exercise 14.3.3 We consider column vectors. Observe that, if x, y ∈ Rn and z = x + iy, then n z 2 = j =1 n = j =1 =x |zj |2 (|xj |2 + |yj |2 ) 2 +y 2 Using the notation introduced in the first paragraph, An z 2 = An (x + iy) 2 = An x + iAn y) = An x 2 2 + An y ) 2 , so An x , An y) → 0 ⇒ An z) → 0. The result follows. 404 Exercise 14.3.7 (i) We have j +s n cij = aik bkj = aik bkj = k=1 k=1 aik bkj = 0 j +s+1≤k≤i−r +1 if i − r ≤ j + s + 1. (ii) If D = (dij ) is diagonal, then D = max |djj | = ρ(D). 405 Exercise 14.3.8 The computations by which we obtained the result for r, s ≥ 0 remain valid and the result remains true. 406 Exercise 14.3.11 Let A be an n × n matrix with non-zero diagonal entries and b ∈ Fn a column vector. Suppose that the equation Ax = b has the solution x∗ . Let us write D for the n × n diagonal matrix with diagonal entries the same as those of A and set B = A − D. If x0 ∈ Rn and xj +1 = D−1 (b − B xj ), then xj − x∗ ≤ D−1 B j and xj − x∗ → 0 whenever ρ(D−1 B ) < 1. If ρ(D−1 ) = 1. we can find an x0 such that xj − x∗ → 0. To prove this, observe that −D−1 B x∗ = D−1 b − x∗ and so xj +1 − x∗ = D−1 (b − B xj ) − x∗ = −D−1 B (xj − x∗ ) Thus xn − x∗ ≤ (D−1 B )n x0 − x∗ If ρ(D−1 B ) < 1 (D−1 B )n → 0 the result follows. If ρ(D−1 B ) ≥ 1 we can find an eigenvector e with eigenvalue having absolute value at least 1. If we set x0 = x∗ + e0 convergence fails. To prove the last part, suppose, if possible, that we can find an eigenvector y of D−1 B with eigenvalue λ such that |λ| ≥ 1. Then D−1 B y = λy and so B y = λDy. Thus n aij yj = λaii yi j =i and so |aii ||yi | ≤ j =i |aij ||yj | for each i and (since y = 0) |aii ||yi | < for at least one value of i. j =i |aij ||yj | 407 Summing over all i and interchanging the order of summation, we get n i=1 n |aii ||yi | ≤ n i=1 j =i n = j =1 |yj | |aij ||yj | = i=j |aij | < j =1 i=j n j =1 |aij ||yj | |yj ||ajj | = n i=1 |aii ||yi | which is absurd. Thus all the eigenvalues of D−1 B have absolute value less than 1 and we may apply the first part of the question. 408 Exercise 14.4.3 If α has matrix A with respect to given orthonormal basis, α∗ has matrix A∗ , αα∗ has matrix AA∗ and α∗ α has matrix A∗ A. Thus αα∗ = α∗ α ⇔ A∗ A = AA∗ . 409 Exercise 14.4.6 If α is diagonalisable with distinct eigenvalues λ1 , λ2 , . . . , λm , then U is the direct sum of the spaces Ej = {e : αe = λj e}. Moreover Ej ⊥ Ek , that is to say ej ∈ Ej , ek ∈ Ek ⇒ ej , ek = 0. when j = k . It follows that if πj is the (unique) orthogonal projection with πj U = Ej we have πj πk = 0 for all j = k If u ∈ U , we can write u uniquely as u = e1 + e2 + . . . + e m ⋆ with ej ∈ Ej . Since πj πk = 0 for all j = k we have πj ek = πj πk ek = 0. Thus, applying πj to both sides of ⋆, πj u = ej . Thus m ιu = πj u j =1 and m αu = m m α ej = λj ej = j =1 j =1 λj πj u. j =1 for all u ∈ U . It follows that ι = π1 + π2 + . . . + πm and α = λ1 π1 + λ2 π2 + . . . + λm πm . Conversely, if the stated conditions hold and we write Ej = πj U , we see that Ej is a subspace and πi Ej = πi πj Ej = 0 for i = j . Since iota = π1 + π2 + . . . + πm , we have m u = ιu = m πj u = j =1 uj j =1 with uj = πj u ∈ Ej . On the other hand, if ej ∈ Ej and m ej = 0 j =1 then applying πi to both sides we get ei = 0 for all i. Thus U = E 1 ⊕ E2 ⊕ . . . ⊕ E m . Moreover, since πj πk = 0 we have Ej ⊥ Ek for k = j . Let Ej be an orthonormal basis for Ej . The set E = m Ej is an j =1 orthonormal basis for U . Since α has a diagonal matrix with respect to this basis, α is normal. 410 α is self-adjoint if and only if there exist distinct λj ∈ R and othogonal projections πj such that πk πj = 0 when k = j , ι = π1 + π2 + . . . + πm textand α = λ1 π1 + λ2 π2 + . . . + λm πm . 411 Exercise 14.4.7 If α is unitary, then α = α−1 so αα∗ = ι = αα∗ and α is normal. β unitary ⇔ ββ ∗ = ι ⇔ α−1 α∗ (α−1 α∗ )∗ = ι ⇔ α−1 α∗ α∗∗ (α∗ )−1 = ι ⇔ α−1 α∗ α(α∗ )−1 = ι ⇔ α∗ α = αα∗ ⇔ α normal 412 Exercise 14.4.9 (i) Choose an orthonormal basis such that α is represented by a diagonal matrix D with djj = eiθj with θj real. Let f (t) be represented by a diagonal matrix D with djj (t) = eiθj t . Then f (t) − f (s) ≤ max |eiθj t − eiθj s | ≤ max |θj ||t − s| j so f is continuous, f (t) is unitary, f (0) = ι, f (1) = α. Recall that the unitary maps form a group. Let α = β −1 γ . Since α is unitary we can find f as in the first paragraph. Set g (t) = βf (t). (ii) By considering matrix representations, we know that det : L(U, U ) → R is continuous. Thus, if f is as stated, det f (t) is a continuous function of t. But det f (1) = −1, det f (0) = 1 so, by the intermediate value theorem, we can find an s ∈ [0, 1] such that det f (s) = 0 and so f (s) is not invertible. 413 Exercise 15.1.8 Observe, that if x = 0, then at least one of the following two things must occur:(A) x1 = 0, so β (x1 , x2 )T , (0, 1)T = x1 = 0. (B) x2 = 0, so β (x1 , x2 )T , (1, 0)T = −x2 = 0. Thus β is non-degenerate, but for all x. β (x, x) = x1 x2 − x2 x1 = 0 We cannot find an α of the type required. If α is degenerate, then there exists an x = 0 with α(x, y) = 0 for all y so, in particular, α(x, x) = 0. 414 Exercise 15.1.13 Observe that, with the notation of Definition 15.1.11, q (u + v) + q (u − v) = α(u + v, u + v) + α(u − v, u − v) = α(u, u + v) + α(v, u + v) + α(u, u − v) + α(v, u − v) = α(u, u) + α(u, v) + α(v, u) + α(v, v) + α(u, u) − α(u, v) − α(v, u) + α(v, v) = 2α(u, u) + 2α(u, u) = 2 q (u) + q (v) . 415 Exercise 15.1.15 Observe that n n n n xi bij xj = i=1 j =1 so n xj bij xi , j =1 i=1 n n n xi bij xj = i=1 j =1 xi i=1 j =1 bij + bji xj 2 and q is a quadratic form with associated symmetric matrix 1 A = 2 (A + AT ). 416 Exercise 15.1.16 {x ∈ R3 : x2 + x2 + x2 = 1} 1 2 3 is a sphere. {x ∈ R3 : x2 + x2 + x2 = 0} 1 2 3 is a point. {x ∈ R3 : x2 + x2 + x2 = −1} = ∅. 1 2 3 {x ∈ R3 : x2 + x2 − x2 = 1} 1 2 3 is a one sheeted hyperboloid of revolution obtained by revolving the hyperbola {x ∈ R3 : x2 − x2 = 1, x2 = 0} 3 1 about its semi-minor axis Ox3 . is a circular cone. {x ∈ R3 : x2 + x2 − x2 = 0} 1 2 3 {x ∈ R3 : x2 + x2 − x2 = −1} 1 2 3 is a two sheeted hyperboloid of revolution obtained by revolving the hyperbola {x ∈ R3 : x2 − x2 = −1, x2 = 0} 1 3 about its semi-major axis Ox3 . (The two sheeted hyperboloid has two connected components. The one sheeted hyperboloid has one.) is a circular cylinder. {x ∈ R3 : x2 + x2 = 1} 1 2 {x ∈ R3 : x2 + x2 = 0} 1 2 is the straight line give along the axis Ox3 . {x ∈ R3 : x2 + x2 = −1} = ∅. 1 2 {x ∈ R3 : x2 − x2 = 1} and x ∈ R3 : x2 − x2 = −1} 1 2 1 2 are hyperbolic cylinders. {x ∈ R3 : x2 − x2 = 0} 2 1 is a pair of planes intersecting at right angles. {x ∈ R3 : x2 = 1} 1 417 is a pair of parallel planes. is a plane. {x ∈ R3 : x2 = 0} 1 {x ∈ R3 : x2 = −1} = ∅. 1 and {x ∈ R3 : 0 = 1} = {x ∈ R3 : 0 = −1} = ∅ {x ∈ R3 : 0 = 0} = R3 . 418 Exercise 15.1.22 By a rotation (which changes neither volume nor determinant) we may suppose that our ellipsoid is n i=1 2 di yi ≤ L. 1/2 1/2 By successive scale changes wi = di yi which multiply volume by di we can transform our ellipsoid to n i=1 2 wi ≤ L Now make a scale change in every direction ui = L−1/2 wi , to obtain the unit sphere of volume Vn Our original volume must have been Ln/2 (det A)−1/2 Ln/2 Vn . n j =1 −1/2 dj Vn i.e, 419 Exercise 15.1.24 We know that 1 fY (y) = K exp − 2 n n 2 dj yj i=1 j =1 is a density function and so fY (y) dV (y) = 1. Rn Thus ∞ 1=K −∞ ∞ 1 ... exp − 2 −∞ −∞ ∞ ∞ =K ∞ −∞ −∞ 2 dj yj dy1 dy2 . . . dyn j =1 n ... n n ∞ ∞ =K j =1 n −∞ −∞ j =1 2 exp(−dj yj /2) dy1 dy2 . . . dyn 2 exp(−dj yj /2) dyj ∞ −1/2 =K dj −∞ j =1 exp(−t2 /2) dtj j n −1/2 = K (2π )n/2 dj . j =1 Thus 1/2 n K = (2π ) −n/2 dj j =1 . 420 Exercise 15.1.25 (i) Suppose α sesquilinear, β Hermitian, γ skew-Hermitian and α = β + γ. Then α(u, v) = β (u, v) + γ (u, v) and α(v, u)∗ = β (v, u) + γ (v, u) Thus ∗ = β (u, v) − γ (v, u). β (u, v) = 1 (α(u, v) + α(v, u)∗ ) 2 1 γ (u, v) = 2 (α(u, v) − α(v, u)∗ ) Conversely, if α is sesquilinear, the definitions 1 β (u, v) = 2 (α(u, v) + α(v, u)∗ ) 1 γ (u, v) = 2 (α(u, v) − α(v, u)∗ ) give sesquilinear forms (the point at issue is that if (u, v) → α(u, v) is sesquilinear, so is the map (u, v) → α(v, u)∗ ) with 1 β (v, u) = 2 (α(v, u) + α(u, v)∗ ) = β (u, v)∗ 1 γ (v, u) = 2 (α(v, u)∗ − α(v, u)) = −γ (v, u)∗ (ii) Since α(u, v) = α(v, u)∗ for all u, v, we have α(u, u) = α(u, u)∗ , so α(u, u) is real for all u ∈ U . (iii) Set β = iα. Then α skew-Hermitian ⇔ α(v, u) = −α(u, v)∗ ∀v, u ⇔ β (u, v)∗ = iα(u, v) ⇔ β Hermitian. ∗ = −iα(u, v)∗ = iα(v, u) = β (v, u) ∀v, u (iv) Observe that α(u + v, u + v) = α(u, u) + α(u, v) + α(v, u) + α(u, u) = α(u, u) + α(u, v) + α(u, v)∗ + α(v, v) = α(u, u) + 2ℜα(u, v) + α(v, v). Thus and α(u − v, u − v) = α(u, u) + 2ℜα(u, v) + α(v, v) α(u + v, u + v) − α(u − v, u − v) = 4ℜα(u, v). It follows that α(u+iv, u+iv)−α(u−iv, u−iv) = 4ℜα(u, iv) = −4ℜ iα(u, v) = ℑ(u, v). The result follows. 421 (v) Choose any orthonormal basis f1 , f2 , . . . , fn for U . Define a matrix A = (apq ) by the formula apq = α(fp , fq ). By sesquilinearity n n n zp fp , α p=1 w q fq n ∗ zp apq wq = q =1 p=1 q =1 for all zp , wq ∈ C. Since α is Hermitian aqp = α(fq , fp ) = α(fp , fq )∗ = a∗ pq Thus A is Hermitian and we can find a unitary matrix M such that M ∗ AM = D where D is a real diagonal matrix whose entries are the eigenvalues of A appearing with the appropriate multiplicities. If we set eq = n=1 mpq fp then (since M is unitary) the e1 , e2 , . . . , en are p an orthonormal basis and by direct calculation n α zr er , r =1 for all zr , ws ∈ C. n n ws e s s=1 ∗ dt zt wt = t=1 422 Exercise 15.1.26 Suppose that U is vector space over C and α : U × U → C is a bilinear form. Let us set θR (w)z = α(z, w) for z, w ∈ U . We observe that θR (w)(λ1 z1 + λ2 z2 ) = α(λ1 z1 + λ2 z2 , w) = λ1 α(z1 , w) + λ2 α(z2 , w) = λ1 θR (w)z1 + λ2 θR (w)z2 for all λ1 , λ2 ∈ C and z1 , z2 ∈ U . Thus θR (w) ∈ U ′ for all w ∈ U . Now θR (λ1 w1 + λ2 w2 )z) = α(z, λ1 w1 + λ2 w2 ) = λ∗ α(z, w1 ) + λ∗ α(z, w2 ) 1 2 = λ∗ θR (w1 )z + λ∗ θR (w2 )z 1 2 = λ∗ θR (w1 ) + λ∗ θR (w2 ) z 1 2 for all z ∈ U . Thus θR (λ1 w1 + λ2 w2 ) = λ∗ θR (w1 ) + λ∗ θR (w2 ) 1 2 and θR : U → U ′ is an anti-linear map. Suppose in addition we know that U is finite dimensional and that α(z, w) = 0 for all z ∈ U ⇒ w = 0. Then θR is injective since θR (w) = 0 ⇒ θR (w)(z) = 0 for all z ∈ U ⇒ α(z, w) = 0 for all w ∈ U ⇒ w = 0. Since dim U = dim U ′ , it follows that θR is an isomorphism. Alternatively, observe that if e1 , e2 , . . . , en form a basis for U n n j =1 λ∗ ej j λ j θR e j = 0 ⇒ θ R =0 j =1 n ⇒ ⇒ λ∗ ej = 0 j j =1 λ∗ = j 0 ∀j ⇒ λj = 0 ∀j so θR e1 , θR e2 , . . . , θR en are linearly independent so a basis for U ′ so span U ′ . Thus θR is surjective. 423 Exercise 15.2.4 Let f1 , f2 , . . . , fn be the basis associated with S . Define n n n α x i fi , i=1 = y j fj j =1 xk yk k=1 for all xi , yj ∈ R. Then, by inspection, α is a symmetric form on Rn . By Theorem 15.2.1 we can find a basis e1 , e2 , . . . , en and positive integers p and m with p + m ≤ n such that n xi ei , α i=1 yj e j n q yi e i i=1 as required. = j =1 for all xi , yj ∈ R. We have p+ m p n k=1 n =α yj e j j =1 xk yk k=p+1 p+ m p n yi e i , i=1 xk yk − 2 yk = k=1 − 2 yk k=p+1 424 Exercise 15.2.8 A is two quadrants, B two lines at right angles. (1, 0), (−1, 1) ∈ A, but (0, 1) = (1, 0) + (−1, 1) ∈ A. Thus A is not / a subspace. (1, 0), (−1, 1) ∈ B , but (0, 1) = (1, 0) + (−1, 1) ∈ B . Thus A is not / a subspace. 425 Exercise 15.2.9 (i) Choose M orthogonal so that M T AM = D with D diagonal. Since matrix rank is unchanged by multiplication on the left or right by invertible matrices, D has the same matrix rank as A. But the signature rank of A and the matrix rank of D are both the number of non-zero entries of D. Thus the signature and matrix ranks are identical. (ii) We continue with the notation of (i). The rank of q is the number of non-zero characteristic roots (multiple roots counted multiply). Now signature q = (no. strictly pos. entries D)−(no. strictly neg. entries D), so the signature of D is the number of strictly positive characteristic roots minus the number of strictly negative characteristic roots of A (multiple roots counted multiply). 426 Exercise 15.2.10 (i) Observe that γ (P z) = (P zT )A(P z) = (zT P T )A(P z) = zT (P T AP )z. (ii) We cannot use results on the diagonalisation of real symmetric matrices or Hermitian matrices, but we can use completing the square. (α ) Suppose that A = (aij )1≤i,j ≤n is a symmetric n × n matrix with a11 = 0, and we set bij = a−1 (a11 aij − a1i a1j ). Then B = (bij )2≤i,j ≤n is 11 a symmetric matrix. Further, if z ∈ Cn and (after choosing one square 1/2 root a11 of a11 ) we set n w1 = 1/2 a11 −1/2 z1 + a1j a11 zj j =2 and wj = zj otherwise, we have n n n zi aij zj = 2 w1 n + i=1 j =1 wi bij wj . i=2 j =2 (β ) Suppose that A = (aij )1≤i,j ≤n is a symmetric n × n matrix. Suppose further that σ : {1, 2, . . . , n} → {1, 2, . . . , n} is a permutation (that is to say σ is a bijection) and we set bij = aσi σj . Then C = (cij )1≤i,j ≤n is a symmetric matrix. Further, if z ∈ Cn and we set wj = xσ(j ) , we have n n n n zi aij zj = i=1 j =1 wi cij wj . i=1 j =1 (γ ) Suppose that n ≥ 2, A = (aij )1≤i,j ≤n is a symmetric n × n matrix and a11 = a22 = 0, but a12 = 0. Then there exists a symmetric n × n matrix C with c11 = 0 such that, if z ∈ Cn and we set w1 = (z1 + z2 )/2, w2 = (z1 − z2 )/2, wj = zj for j ≥ 3, we have n n n n zi aij zj = i=1 j =1 wi cij wj . i=1 j =1 Combining (α), (β ) and (γ ) we have the following. (δ ) Suppose that A = (aij )1≤i,j ≤n is a non-zero symmetric n × n matrix. Then we can find an n × n invertible matrix M = (mij )1≤i,j ≤n and a symmetric (n − 1) × (n − 1) matrix B = (bij )2≤i,j ≤n such that, if z ∈ Cn and we set wi = n=1 mij zj , then j n n n zi aij zj = i=1 j =1 2 w1 n + wi bij wj . i=2 j =2 427 Repeated use of (δ ) gives the desired result. (iii) Observe that if P is invertible so is P T , so (using matrix rank) rank(P T AP ) = rank A. Thus, if r 2 zu , γ (P z) = u=1 we have r = rank A. (iv) Observe that γ (iz) = −γ (z). (v) By (i) and (ii) we can choose a basis ej for Cn so that r n γ zj ej j =1 2 zu . = u=1 Let s be the integer part of r/2. If F is the subspace spanned by the vectors e2k−1 − ie2k with 1 ≤ k ≤ s and the vectors el with r + 1 ≤ l ≤ n. Then F has dimension at least m and any subspace E of F with dimension m will have desired properties. (vi) Let E0 = Cm . Using (v) and proceeding inductively we can find subspaces Ej such that Ej is a subspace of Ej −1 , dim Ej ≥ [2−j n] (using the standard integer part notation) and γj |Ej = 0 [1 ≤ j ≤ k ]. We have dim Ek ≥ 1, so we may choose a non-zero z ∈ Ek to obtain the required result. 428 Exercise 15.2.12 If a = 0, we have Example 15.2.11. The rank is 3 and the signature −1. If a = 0, then, setting y1 = (x1 − 2−1 a−1 x2 − 2−1 a−1 ), y2 = x2 , y3 = x3 , we have 2 2 2 x1 x2 + x2 x3 + x3 x1 + ax2 = ay1 +4−1 a−1 y2 +4−1 a−1 y2 +(1 − 2−1 a−1 )y2 y3 1 Setting u1 = y1 , u2 = y2 + 2a(1 − 2−1 a−1 )y3 u3 = y3 , we have x1 x2 + x2 x3 + x3 x1 + ax2 = au2 + 4−1 a−1 u2 − a(1 − 2−1 a−1 )2 u3 . 1 1 2 If a = 1/2, we see that q has rank and signature 2. If a > 0 and a = 1/2, q has rank 3 and signature 1. If a < 0, q has rank 3 and signature −1. 429 Exercise 15.2.13 (i) We know that we can find a unitary matrix U such that C = U AU is diagonal with real entries. Let D be the diagonal matrix with −1/2 (positive square root) otherwise. dii = 1 if cii = 0 and dii = cii Setting P = DU we see that P ∗ AP is diagonal with diagonal entries taking the values 1, −1 or 0. ∗ (ii) We have n ∗ n ∗ zr ars zs n n ∗ zr a∗ zs rs = r =1 s=1 r =1 s=1 n n ∗ zr asr zs = r =1 s=1 n n ∗ zr ars zs = r =1 s=1 so n r =1 n ∗ s=1 zr ars zs is real. ∗ (iii) If P1 AP1 has k entries 1, then if E is the subspace of Cn spanned by P1 e with e a column vector with ith entry 1 and all other entries 0 ∗ when the ith diagonal entry of P1 AP1 is 1, we have dim E = k, z∗ Az > 0 ∀z ∈ E \ {0}. ∗ If P2 AP2 has k ′ entries 1 then if F is the subspace of Cn spanned by P2 e with e a column vector with ith entry 1 and all other entries 0 ∗ when the ith diagonal entry of P1 AP1 is −1 or 0, we have dim F = n − k ′ , z∗ Az ≤ 0 ∀z ∈ F. Thus E ∩ F = {0}, dim E + dim F ≤ n and n − k ′ + k ≤ n. It follows that k ≤ k ′ . The same argument shows that k ′ ≤ k so k = k ′ . ∗ In the same way (or by considering −A) we see that P1 AP1 and have the same number of entries −1 and so since they have the same number of entries 1 and −1 must have the same number of entries 0 along the diagonal. ∗ P2 AP2 430 Exercise 15.3.2 (i) Let U be a vector space over R. A quadratic form q : U → R is said to be negative semi-definite if q (u) ≤ 0 for all u ∈ U and strictly negative definite if q (u) < 0 for all u ∈ U with u = 0. By inspection q : U → R is strictly negative definite (respectively negative semi-definite) if and only if −q is strictly positive definite (respectively positive semi-definite). (ii) If q is a quadratic form over a finite dimensional real vector space, then we can find a basis e1 , e2 , . . . , en such that n r xj ej q j =1 = j =1 r +s x2 − j x2 . j j =r +1 Setting r n xj ej q1 j =1 = j =1 r +s n x2 j xj ej and q2 j =1 =− x2 j j =r +1 we see that q = q1 + q2 with q1 a positive semi-definite quadratic form and q2 a negative semi-definite quadratic form. The observation 0 = 0 + 0 = x 2 − x2 shows that the decomposition is not unique even for a space of dimension 1. 431 Exercise 15.3.3 We have n n 2 n ci (EXi Xj )cj = E i=1 j =1 cj Xj j =1 with equality if and only if n Pr cj Xj = 0 j =1 = 1. ≥0 432 Exercise 15.3.4 We know that we can find a basis ej such that n u xj e j α j =1 v x2 j = j =1 − x2 . j j =u+1 If v < n, then q (en ) = 0 and q is not strictly positive definite. If v ≥ u + 1, q (eu+1 ) = −1 so q is not positive semi-definite. By inspection, u = v = n implies q strictly positive definite so q is strictly positive definite if and only if n = u = V i.e. q has rank and signature n. By inspection, u = v implies q positive semi-definite so q is positive semi-definite if and only if u = v , i.e. q has rank and signature equal. Finally q is negative semi-definite if and only if −q is positive semidefinite, i.e. q has signature equal to minus its rank. 433 Exercise 15.3.5 The conditions (i) α(u, v) = α(v, u) (ii) α(λu, v) = λα(u, v) (iii) α(u + w, v) = α(u, v) + α(w, v) (for all u, v, w ∈ U and λ ∈ R) say that α is a quadratic form. The additional conditions (iv) α(u, u) ≥ 0 ∀u ∈ U (v) α(u, u) = 0 ⇒ u = 0 say that α gives rise to a positive definite quadratic form. Conditions (i) to (v) say that α is an inner product. 434 Exercise 15.3.6 By translation we may take a = 0 and by subtracting a constant we may take f (0) = 0. Since f is smooth we know by the local Taylor’s theorem that ∂f ∂f f (h) = (0)hi + (0)hi hj + ǫ(h) h 2 ∂xi ∂xi ∂xj with ǫ(h) → 0 as h → 0. (We use the summation convention.) (i) Suppose we have a minimum at 0. If ei is the vector with 1 in the ith place and 0 elsewhere ∂f f (hei ) (0) = lim ≥0 hi →0+ ∂xi h and ∂f f (hei ) ≤0 (0) = lim hi →0− ∂xi h ∂f so ∂xi = 0. We now have f (h) = Hij hi hj + ǫ(h) h 2 with H the Hessian. Let u be any non-zero vector We have 0 ≤ f (hu) = h2 Hij ui uj + ǫ(hu) u so and, allowing h → 0, 0 ≤ Hij ui uj + ǫ(hu) u 2 , 2 0 ≤ Hij ui uj . Thus H is positive semi-definite. (ii) We have f (h) = Hij hi hj + ǫ(h) h 2 with H the Hessian. Since u → H u is continuous and the surface S of the unit ball is compact, we know that H takes a minimum on S which is non-zero so there exists an η > 0 with for all u = 1. H (u) ≥ η Thus f (h) ≥ h 2 (η − ǫ(h) . There exists a δ > 0 such that |ǫ(h)| < η/2 for all h < δ . If h < δ , then f (h) ≥ (η/2) h 2 . Thus f has a strict minimum at 0. 435 (ii) Define f : R → R by f (x) = x4 . We have a strict minimum at 0, but the Hessian H = f ′′ (0) = (0) is not strictly positive definite. 436 Exercise 15.3.7 Since ∂f ∂f (x, y ), (x, y ) ∂x ∂y the stationary points are = (cos x sin y, sin x cos y ) 1 1 (x, y ) = nπ, mπ and(x, y ) = (n + 2 )π, (m + 2 )π . The Hessian H= − sin x sin y cos x cos y cos x cos y − sin x sin y so if (x, y ) = nπ, mπ H= 0 (−1)n+m (−1)n+m 0 which is neither positive semi-definite nor negative semi-definite (note for example that 4xy = (x + y )2 − (x − y )2 ) and we have a saddle, 1 whilst if (x, y ) = (n + 2 )π, (m + 1 )π 2 H= (−1)n+m 0 0 (−1)n+m which is strictly positive definite if n+m is even (so we have a minimum) and strictly negative definite if n + m is even (so we have a maximum). 437 Exercise 15.3.11 We have ˜ ˜˜ ˜˜ ˜˜ ˜ AT = (LLT )T = LT T LT = LLT = A, ˜ so A is symmetric. Observe that ˜ ˜˜ ˜ xT Ax = xT LLT x = LT x 2 ≥0 ˜ ˜ so A is symmetric positive semi-definite. The case L = 0 shows that A need not be strictly positive definite. If L is lower triangular with non-zero diagonal entries, then det L = 0, so L is non-singular and xT Ax = 0 ⇒ xT LLT x = 0 ⇒ LT x 2 =0 T ⇒ L x =0 ⇒ LT x = 0 ⇒ x = 0, so A = LLT is a symmetric strictly positive definite matrix. 438 Exercise 15.3.12 If, at the first stage a11 < 0, then A is not positive definite. If a11 > 0, then either the Schur matrix B is positive definite, in which case, A is or the Schur matrix B is not positive definite, in which case, A is not. (Consider vectors whose first term is 0.) If at each stage, the upper corner entry is strictly positive then our proof shows that A is positive definite. If at some stage it is not, we know that A is not positive definite. 439 Exercise 15.3.14 We need to look 4 T −6 A − l1 l1 = 2 The Schur matrix at l1 = (2, −3, 1)T . Now 00 0 4 −6 2 −6 2 8 −5 − −6 9 −3 = 0 −1 −2 . 0 −2 13 2 −3 1 −5 14 −1 −2 −2 13 has strictly negative upper corner entry, so A is not positive semidefinite. B= 440 Exercise 15.3.15 (i) Look at the first formula of the proof of part (ii) of Theorem15.3.10. (ii) Looking at the proof of of part (i) of Theorem15.3.10 we see that the computation of l involves roughly n operations and the computation of B involves roughly n2 operations. Thus the reduction of our problem from n × n matrices to (n − 1) × (n − 1) matrices involves roughly 2n2 operations and the total number of operations required is roughly n 2 2 r 2 ≈ n3 3 r =1 operations. [If the reader’s definition of an operation agrees with mine, this argument shows that we can certainly take K = 1 (and, if we really needed, as close to 2/3 as we wished) for n large.] 441 Exercise 15.3.16 11 If we take l1 = (1, 2 , 3 )T , then 1 T 1 H3 − l1 l1 = 2 1 2 1 3 1 4 1 3 0 0 = 0 1 1 If we take l2 = ( 12 )1/2 , ( 12 )1/2 1 12 1 12 1 12 4 45 0 1 3 1 4 1 5 0 1 2 1 4 1 6 1 1 − 2 1 3 1 3 1 6 1 9 1 12 4 45 1 12 1 12 T , then 0 0 − l2 lT = 2 0 1 180 Thus H3 = LLT with 1 0 0 0 L = 2−1 12−1/2 −1 −1/2 3 12 180−1/2 Second part 111 If we take l1 = (1, 2 , 3 , 4 )T , then 11 123 1 1 1 2 H4 − l1 lT = 1 3 4 1 11 3 1 4 4 1 5 5 1 6 0 0 = 0 0 0 0 1 12 1 12 3 40 1 12 4 45 1 12 1 4 1 5 1 6 λ 1 2 1 4 1 6 1 8 1 1 − 2 1 3 1 4 0 3 40 1 12 λ− 1 3 1 6 1 9 1 12 1 16 1 4 1 8 1 12 1 16 3 If we take l2 = ((12)−1/2 , (12)−1/2 , (12)1/2 40 )T then 1 1 1 1 1 3 3 12 1 12 3 40 12 4 45 1 12 40 1 12 1 16 − l2 lT = 2 λ− 0 0 0 = 0 180−1 120−1 13 0 120−1 λ − 100 12 1 12 3 40 12 4 45 1 12 40 1 12 λ− 1 16 − 12 1 12 3 40 1 12 1 12 3 40 3 40 3 40 27 400 442 If we take l3 = ((180)−1/2 , (180)1/2 120−1 )T ) 180−1 120−1 13 120−1 λ − 100 − l3 lT = 3 0 0 57 0 λ − 400 Thus λ0 = 57/400. We observe that 1 1 . λ0 − = 7 2800 (The ‘Hilbert matrices’ Hn are ‘only just positive definite’ and ‘only just invertible’ and are often used to test computational methods.) If λ ≥ λ0 then H4 = LLT with 1 0 0 0 2−1 12−1/2 0 0 L = −1 3 12−1/2 180−1/2 0 57 1/2 −1 1/2 3 1/2 −1 4 (12) 40 (180) 120 (λ − 400 ) 443 Exercise 15.3.17 (i) The result is true when n = 1, since then the only root is a0 which is positive by hypothesis. Suppose the result is true for m ≥ n ≥ 1 and P is a polynomial of the given form of degree m + 1. We observe that P ′ is a polynomial of the given form of degree m and so all its real zeros are strictly positive. Thus, if m + 1 is even, P ′ (t) < 0 for t ≤ 0, so P is decreasing as t runs from −∞ to 0. But P (0) = a0 > 0 so P (t) > 0 for t ≤ 0 and P has no non-positive real zeros. If m + 1 is odd, P ′ (t) > 0 for t ≤ 0, so P is increasing as t runs from −∞ to 0. But P (0) = −a0 < 0, so P (t) < 0 for t ≤ 0 and P has no non-positive real zeros. The result follows by induction. (ii) The only if part is immediate. To obtain the if part apply part (i) to n P (t) = det(tI − A) = j =1 n (t − λj ) = tn + j =1 (−1)n−j aj tj , observing that n an−1 = j =1 λi λj λk . . . . . λi λj > 0, an−3 = λj , an−2 = i,j,k distinct j =i (iii) Let A= 1 −3 . 31 Then det(tI − A) = (t − 1)2 + 9 = t2 − 2t + 8 is a polynomial with no real roots (since (t − 1)2 + 9 > 0 for all real t). (iv) Observe that with det(tI − A) = t3 − b2 t2 + b1 t − b0 b2 = a11 + a22 + a33 b1 = a11 a22 + a22 a33 + a33 a22 − a12 a21 − a23 a32 − a31 a13 > 0 b0 = det A and apply the results above. 444 Exercise 15.3.19 Suppose P is invertible with P T AP and P T BP diagonal. Since A is not semi-positive definite P T AP is not and, since A is not semi-negative definite, P T AP is not. Thus a0 0b P T AP = with a and b non-zero of opposite sides. Observing that 01 10 c0 0d 01 10 = c0 0d and λ2 c 0 λ0 c0 λ0 = 0 µ2 d 0µ 0d 0µ we see that there is a non-singular matrix Q with a0 Q=A 0b QT and QT CQ diagonal whenever C is diagonal. Setting M = P Q we see that M is invertible M T AM = A and M T BM is diagonal. Let ab . cd M= Since M T AM = A we have 10 = A = M T AM 0 −1 a b −c −d = ac bd = a2 − c2 ab − cd ba − cd b2 − d2 Thus ab = cd and d > 0, a > 0 On the other hand M T BM = ac bd cd ab = 2ac ad + bc bc + ad 2bd so if M T BM is diagonal bc = −ad. We now get c2 d = abc = −a2 d so, since d = 0, we have c = a = 0 which contradicts the condition a2 − c2 = 1. 445 (ii) Since det A, det B = 0, A and B have rank 2. A has signature 0 1 by inspection. B has signature 0 since xy = 4 (x + y )2 − (x − y )2 . The transformation x → M x leaves lines x = y and x = −y unchanged (or interchanges them) since these are the asymptotes to x2 − y 2 = K for all K . It therefore gives dilation along the directions x = y x = −y (and possibly interchanges these two axes) and will therefore not transform xy = k in the desired manner. (iii) C is strictly negative definite and the same argument which gave simultaneous diagonalisation when one form is strictly positive definite will work if one form is strictly negative definite. 446 Exercise 15.4.2 We have n n q xi e i =α i=1 n = n xi ei , i=1 n xj ej j =1 xi aij xj i=1 j =1 n n = 1 2 (xi aij xj + xi aji xj ) = i=1 j =1 1 2 n n 0 = 0. i=1 j =1 447 Exercise 15.4.3 Observe that (iA)∗ = −iA∗ = −iAT = iA. Thus iA is Hermitian and so has real eigenvalues. If −λ is an eigenvalue of iA then iλ = (−i)(−λ) is an eigenvalue of A = (−i)(iA). Thus the eigenvalues of A are purely imaginary. Observe that (M T BM )T = M T B T M T T = −M T BM so the eigenvalues of M T BM are purely imaginary. But M T BM is a real diagonal matrix whose diagonal entries are its eigenvalues which are thus real and so must be zero. Thus M T BM = 0 and so B = 0. 448 Exercise 15.4.6 (i) Let uj [1 ≤ j ≤ n] form a basis for an n-dimensional complex vector space U . Let α be the antisymmetric form associated with A for this basis. Let ej [1 ≤ j ≤ n] be the basis found in Theorem 15.4.5. If M is the appropriate basis change matrix M T AM has the required form. (ii) Observe that 2m = rank M T AM = rank A. 449 Exercise 15.4.7 Observe that, by Theorem 15.4.5, if α is non-singular, 2m = n, so n is even. 450 Exercise 15.4.8 (i) True. Since we can find a non-singular M such that M T AM has m copies of 01 −1 0 along the diagonal and all other entries 0 rank A = rank M T AM = 2m. (ii) False. Take A= 02 . −2 0 (iii) True. The non-real roots of a real polynomial occur in conjugate pairs. PA is a real polynomial with roots the eigenvalues (over C) and these are purely imaginary (see Exercise 15.4.3). Thus m n−2m PA (t) = t j =1 m (t + idr )(t − idr ) = (t2 + d2 ) r j =1 with dr real. (iv) False. Take A= 04 . −1 0 (v) True. Let A be the n × n matrix with the m matrices 0 dr −dr 0 along the diagonal and all other entries 0. 451 Exercise 15.4.9 We observe that A is skew-Hermitian ⇔ A = iB where B is Hermitian. (i) If A is skew-Hermitian, then, since iA is Hermitian, we can find unitary P such that iP ∗ AP = P ∗ (iA)P is a real diagonal matrix D and so P ∗ AP = −iD is diagonal with diagonal entries purely imaginary. (ii) If A is skew-Hermitian, then, since iA is Hermitian, Exercise 15.2.13 tells us that we can an invertible matrix P such that P ∗ iAP such that P ∗ iAP is diagonal with diagonal entries taking the values 1, −1 or 0 and so P ∗ AP is diagonal with diagonal entries taking the values i, −i or 0. (iii) Suppose that A is a skew-Hermitian matrix and P1 , P2 are invert∗ ∗ ible matrices such that P1 AP1 and P2 AP2 are diagonal with diagonal entries taking the values i, −i or 0. Then the number of entries of each type is the same for both diagonal matrices. This follows at once from Exercise 15.2.13 (iii) by considering iA. ...
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This note was uploaded on 02/18/2012 for the course MATH 533 taught by Professor Drewarmstrong during the Fall '11 term at University of Miami.

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