601-hw-weeks6-7

601-hw-weeks6-7 - Exercises 4.7 p 336 1 l 1 1 I A is Syntmetric so A is diagonnlizable For 5 z 33 A Z 2 u E_I 1" = —2 l i[1‘1 S“AS=[1

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Exercises 4.7, p. 336 1 l 1 1 I. A is Syntmetric, so A is diagonnlizable. For 5' z 33- A Z 2: u : E ._I 1" = —2 l i [1‘1] S“AS=[1 0]ands-'A59= ' 2:2 1 1 ' 0 3 ‘ Q=[U.v].QTAQ=[ ] 0 2 l: l 0 ];therefore A5 2|: 122 "121 l 1 _ 1 l o 243 —121 122 35.A=1,u=E l ,v_fi _l , 3. A is not diagonalizable; A = —l is the only eigenvalue —[ JQTAQ— 10 Q_"’v’ ” 0 3 1 and x = a l: l :| , a 7’: O, are the only eigenvectors. 5. A is diagonalizable Since A has distinct eigenvalues. For Excrmses 43’ p' 349 1 0 1 0 [ 4 ] [ 2 ] .— 1. = , I I = , S‘lAS = ’ |: 10 I :i |: 0 2 ] 2 4 4 2 1 0 X3 2 9 X4 = and 34/155 = ; therefore 2 4 O 32 80 68 1 0 3. = 1 2 } A5 = . 112 E24 310 32 . . . . 6 64.25 7. A is not diagonahzable; A = l 15 the only eigenvalue and x3 = 9 , x4 : ithas geometric multiplicity 2. Abasis for the eigenspace 1" 12175 consists of [1, l, 0]T and[2,0,1]T. 9. A is diagonalizable since A has distinct eigen~ 5‘ X' = |: —3 H1 1 43 “9 values. FOrS = 1 l 2 , X3 = 19 ’ x4 = 62 —7" —2 2 l A, —l 1 0 0 7.xk=3(1)k I +(—l) 1 = S"AS = 0 2 0 .Therefore 3 + (—I)“' 2 3 + (—1)* 4 163 -—ll —?1 2 A5 = _1?2 10 75 _ X“) = |: 4 :’ ; the SEQUCHCE {Xi} has no limit, 324 —22 —141 but “Xi-ll E ‘55. 11. A is not diagonalizable; A = 1 has algebraic multiplicity 1 _1 2 and geometric multiplicity 1. 9. x3. : 661(1)" I: :I — 640/4)" ’: :I = l 13. Q is orthogonal. _ I l + 0/4)" 64.25 15. Q15 not orthogonal since the columns are not 64 k ; X4 2 , orthonormal. 2 _ (1/4) 12135 17. Q is orthogonal. 64-00006 h { } x10 2 ; t e sequence x con- 19...: : 1N5, fl = 1N5, a = 2mg, 5 = 1N; 12199994 * c : lXx/g verges to [64, 1281T. _2 ] 7. Not a vector space 9. Not a vector space 2 11.11;. = (3/4)(3)*[ 1 ] +(5/4)(—1)"[ 1 11. Not a vector space 13. A vector space 1 6(3)k _ 10(_1)k 119 15. Not a vector space 25. A vector space 3 [ 3(3):: + 5(__1)k :] ' x“ = [ 62 ] ; 27. Not a vector space 29. A vector space 88571 1 _ X") = ; the sequence {xk} has no l1m1t . 44288 Exeruses 5.3, p. 373 and "ka _> 00. 1. Not a subspace 3. A subspace _3 _1 5. A subspace 7. Not a subspace 13. xk = —2(1)* 1 — 2a)" 1 — 9. A subspace 11. Not a subspace —7 —2 13. A subspace 15. Not a subspace k k 1 6 + 2(2) — 50—1) 17- PU) = ‘P1(x) + 3172(1)‘ 2173(1) 5H)" 2 = -2—2(2)" —10(—1)* ; 19. A = (—1 ~2x)Bl +(2+3x)132 +xB3 —3B4, 2 14 + 4(2)* — 10(- 11* x arbitraw 33 2049 21. cos 2x = — sin2 x + cos2 x x4 = —44 ; xlo = —2060 ; 23- W = Sp{1.x2} 1 1 0 68 4100 25. In Exercise 2, W = Sp J , the sequence {x1} has no limit and llxkll —> oo. 0 0 0 5 322 2'1 —201 000 000 1. =1 _— 9 a v xme1 81 000 100 010 O ’2 1 o 0 0 l7.x(1)= —2 —e2’ —3 +e3’ 2 |: 0 0 l ]};in Exercise 3,W = 2 l O 16 3 SH—l-l 1:][000] __ . =_ k p 3 y 21.a_ .18,x;t 118(l) [10]+ 0 O 0 1 0 0 O 0 0 7 ( my: 10 . melimitis 16 3 l: 0 1 0 ; in Exercise 5, W = 118 ' —6 ’ 118 10 ' Sp{—-l + .r, —2 + x2}; in Exercise 6, W = Sp{l, -4x + x2}; in Exercise 8, CHAPTERS W =Sp{l —x2,x} 27- W = SP{31.32, E12. 513.521. 523. 531. 532}. Exercises 5.2, p. 366 1. 0 —7 5 12 ~22 38 -1 0 0 —11 _3 -12 ' _50 _6 _,15 ‘ whereBl= 0 1 0 and 0 0 0 7 —21 28 —42 —7 —14 —l 0 0 3. eX—Zsinx,e‘—231nx+3.1x2+1’_2ex_sinx+ 32: 0 0 O 3\/.\'2 +1 0 O 1 5- CI = —2 +03. 62 = 3 —- C3, 63 arbitrary 29. A = B+thereB = (A+AT)/23ndC = (A—AT)/2 19. 0011 21. —4—20 1010 1330 0100 —'12—3 0003 23. 200 0—30 003 31. T(ao + alx + azx2 + a3x3) = (a0 + 2a2) + (a1 + as)x + (—ao + an - as)x2 Exercises 5.10, p. 438 l 0 1- T011) = “1. 7012) = 302.[ 0 3 :I 3. T(A.) = 2A., T(A2) = —2A2. T(A3) = 3/13, 2 0 0 0 0—2 0 0 T(A4)=-3A4, 0 0 3 0 0 0 0—3 5. 1-1—1 1—1 0 ; —1 2 l p(x) = (1 +x —x2)+(1+x2); (10:) = -5(1 +x - x2) — 3(1 +x2) +7(1 + x); s(x) = —2(1+x—x2)—(l +x2)+2(l+x); r(x) = (00 — a1 — a2)(l +x —x2) + (00 — 01)(1+X2) + (—00 + 2a1+a2)(1+ x) 7. 1/3 5/3 [1/3—1/3 9. —1 1 2 3 1 0 0—3 0 0 1 0 o 0 0 1 p(x) = —7x + 2(x + I) + (x2 — 2r); q(x)=13x — 4(x + 1) + (x3 + 3); r(x) = -7x + 3(x + l) — 2(x2 — 2x) + (X3 + 3) 11. The matrix of T with reSpect to B is 2 1 Q1 = I: l 2 :l . The transition matrix from C to —l l B is P = [ l l The matrix ofT with respect 1 0 toCisQ2,whereQ2=P"QlP=[0 3]. 13. The matrix of T with respect to B is —3 0 0 5 Q] = 0 3 —5 0 .This transition 0 0—2 0 0 0 O 2 1 0 0 1 matrixfromCtoBisP: 0 I l 0 .The . 0 l 0 0 l 0 0 0 matrix of T with respect to C is Q2, where Q2 = 2 0 0 O P_IQIP= 0—2 0 0 0 O 3 0 O 0 0-3 I l 0 15.a)Q= 0 2 4 z 0 0 3 l l 2 l O O b)S= 014 ;R= O 2 0 . O 0 1 0 0 3 c)C={l,l+x,2+4x+x2}; 1—-l 2 d)P= 0 1—4 ; O 0 1 e) T(w.) = —1 + 18x +3x2; T(w2) = 5 +41: + 3x2; T(w3)=1+2x+6x2 CHAPTER 6 Exercises 6.2, p. 453 —2 l.—5 3.0,x=[ I] 5. 25 7. 6 9- A11: “2. A12 =6.A13 = ‘2, A33 =1 11- 1411= ’2. A12 =74A13 = -3. A33 = 3 l3. A11=3. A12 = —6. A13 = 2.4433 = —3 15. 8 17. -35 19. —11 21. —9 23. 22 10 5 —10 29. C = -5 -—l 4 —5 —3 7 35. a) H(n) = n!/2; b) 3 seconds forn = 2; 180 seconds for n = 5; 5,443,200 seconds for n = 10 Exercises 6.3, p. 463 1.121 100 —4—1 2 01: 2—4—1 = l—30 1—11 1—30 =~3 3-012 102 100 312=_132=_130 203 023 023 30 =— =—9 273 5-013 103 100 212=_122=_12—1 112 112 11—1 2—1 =— :1 1—1 7.—6 9.3 11.3 13. Use the column interchanges: is 6. 15. Use the column interchanges: [C21 Cl! C39 (241—) [C21 C4! C39 C11.) [C2, C4. C1, C3]; the determinant is —12. [C1, C2. C3. C4] _> [C1, C4, C3, C2] —> [C1, C4, C2, C3]; the determinant [C17 C2! C3! C4] —> 17. 2 4—2—2 2000 1312 1123 1313=‘1124 —1212 -1401 123 100 =2124=2101 401 4—8—11 =201=16 —8—ll 19.1203 1203 2511 011—5 2043:0—4 4—3 0162 0162 1—5 11—5 =—44—3=08—23 162 057 8—23 =IS7=171 21.1121 1121 0141 0141 2130:0—1—1—2 2212 00-30 141 141 = —1—1—2=03—1 0—30 0—30 3—1 =_30=—3 Exercises 6.4, p. 470 l 0 3 1 0 0 l.A—> 2 l l ——> 2 1 —5 0 4 l 4 3—11 3 J ;det(A) = —4. A00 1 2 1 4 3 —> ...
View Full Document

This note was uploaded on 04/18/2009 for the course MATH 601 taught by Professor Un during the Fall '08 term at Ohio State.

Page1 / 4

601-hw-weeks6-7 - Exercises 4.7 p 336 1 l 1 1 I A is Syntmetric so A is diagonnlizable For 5 z 33 A Z 2 u E_I 1" = —2 l i[1‘1 S“AS=[1

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online