This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 email 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
nonzero. 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 nonzero 1111
1111
1 2 2 2 and 1 1 1 1 .
1111
1111 Not possible. All elementary operations (which change a matrix)
involve a nonzero row or column which remains nonzero (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
= BA2 sin2 θ + (BC  − BA cos θ)2 = BA2 (cos2 θ + sin2 θ) + BC 2 − 2BABC  cos θ = BC 2 + BA2 − 2BC  × 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 nontrivial 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 nonzero 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 − 10t), 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 ranknullity 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 nonsingular 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 premultiplication replacing postmultiplication 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 nonzero 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 nonsingular, 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 nonsingular (for example, because the determinant is nonzero), 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 nonsingular (for example, because the determinant is nonzero) 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 nonsingular matrix (e.g. by looking at the determinant), so
r = 4 and the original matrix was nonsingular. 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 nonzero. 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 nonsingular P .
I −1 AI = IAI = A so A ∼ A. If A ∼ B then B = P −1 AP for some nonsingular P and writing
Q = P −1 we have Q nonsingular 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 nonsingular 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 nonzero vector is an
eigenvector with eigenvalue 1. If θ ≡ π mod 2π , then ρθ = −ι and
every nonzero 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 nonzero 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 nonsingular, 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 nonzero, 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 = (GP 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 nonzero 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 nonsingular U is, so all its diagonal entries are nonzero. 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 nonsingular 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 nonsingular 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 GrammSchmidt 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 nonzero 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 nonzero 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 α2 .
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 αe⊥ : e⊥ → e⊥ to be the restriction of α to e⊥ .
1
1
1
1
We observe that αe⊥ is symmetric and e⊥ has dimension m so, by the
1
1
inductive hypothesis, we can ﬁnd m orthonormal eigenvectors of αe⊥
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 LeviCivita 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 nonzero 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 αM : M → M is surjective so invertible so αm is so x = 0. By
M
the ranknullity 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 θU = ιU and since θV = −ιV .
(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 αU = ιU .
If u ∈ W , then u = (ι − α)v for some v ∈ V , so
so αW = 0W 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 = αU u = αu = αW 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 nontrivial 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 nonempty 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 nonzero 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 nonsingular 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 ranknullity 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 nonzero 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 nonzero 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 ranknullity 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∗∗
antiisomorphism. 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 λaij  = 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 λaij  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  = λλ−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 nonzero 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 selfadjoint 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 nondegenerate, 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 semiminor 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 semimajor 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 skewHermitian 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 skewHermitian ⇔ 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 antilinear 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 nonzero 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 nonzero 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 nonzero 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 nonzero 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 semideﬁnite 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 semideﬁnite) if and only if −q is strictly positive deﬁnite
(respectively positive semideﬁnite).
(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 semideﬁnite quadratic form
and q2 a negative semideﬁnite 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 semideﬁnite.
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 semideﬁnite so q is positive
semideﬁnite if and only if u = v , i.e. q has rank and signature equal.
Finally q is negative semideﬁnite 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 nonzero 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 semideﬁnite.
(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 nonzero 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 semideﬁnite nor negative semideﬁnite (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 semideﬁnite. The case L = 0 shows that A
need not be strictly positive deﬁnite.
If L is lower triangular with nonzero diagonal entries, then det L =
0, so L is nonsingular 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 nonpositive 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 nonpositive 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 semipositive deﬁnite P T AP is not and, since A is not seminegative
deﬁnite, P T AP is not. Thus
a0
0b P T AP = with a and b nonzero 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 nonsingular 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 ndimensional 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 nonsingular, 2m = n, so n
is even. 450 Exercise 15.4.8
(i) True. Since we can ﬁnd a nonsingular 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 nonreal 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 skewHermitian ⇔ A = iB where B is Hermitian.
(i) If A is skewHermitian, 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 skewHermitian, 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 skewHermitian 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. ...
View
Full
Document
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.
 Fall '11
 DrewArmstrong
 Vectors, The Land

Click to edit the document details