Introduction to Linear algebra-Strang-Solutions-Manual_ver13

22 a d 1 1 2 b d 1 1 0 c d 1 1 1 these are not

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Unformatted text preview: orthogonal vecp tors are independent. (c) Starting p from q 1 D .1; 1; 1/= 3 my favorite is q 2 D p .1; 1; 0/= 2 and q 3 D .1; 1; 2/= b . 5 Orthogonal vectors are .1; 1; 0/ and .1; 1; 1/. 1 1 .p ; p ; 3 3 1 p /. 3 1 Orthonormal are . p ; 2 1 p ; 0/, 2 # 4 =9 2=9 . 5=9 Solutions to Exercises 49 TT T 6 Q1 Q2 is orthogonal because .Q1 Q2 /T Q1 Q2 D Q2 Q1 Q1 Q2 D Q2 Q2 D I . 7 When Gram-Schmidt gives Q with orthonormal columns, QT Qb D QT b becomes x b D QT b. x 8 If q 1 and q 2 are orthonormal vectors in R5 then .q T b/q 1 C .q T b/q 2 is closest to b. 1 2 9 10 11 12 13 14 " # " # :8 :6 100 T :8 has P D QQ D 0 1 0 (a) Q D :6 (b) .QQT/.QQT / D 0 0 000 Q.QTQ/QT D QQT . (a) If q 1 , q 2 , q 3 are orthonormal then the dot product of q 1 with c1 q 1 C c2 q 2 C c3 q 3 D 0 gives c1 D 0. Similarly c2 D c3 D 0. Independent q ’s (b) Qx D 0 ) Q T Q x D 0 ) x D 0. 1 1 (a) Two orthonormal vectors are q 1 D 10 .1; 3; 4; 5; 7/ and q 2 D 10 . 7; 3; 4; 5; 1/ (b) Closest in the plane: project QQT .1; 0; 0; 0; 0/ D .0:5; 0:18; 0:24; 0:4; 0/. (a) Orthonormal a’s: aT b D aT .x1 a1 C x2 a2 C x3 a3 / D x1 .aT a1 / D x1 1 1 1 (b) Orthogonal a’s: aT b D aT .x1 a1 C x2 a2 C x3 a3 / D x1 .aT a1 /. Therefore x1 D 1 1 1 a T b =a T a 1 1 1 (c) x1 is the first component of A 1 times b. T T The multiple to subtract is aT b . Then B D b aT b a D .4; 0/ 2  .1; 1/ D .2; 2/. aa aa p  p p    p   kak q T b 14 1=p2 1=p2 2 2p2 1 D q1 q2 D D QR. 10 0 kB k 1= 2 1= 2 022 1 .1; 2; 3 1 2/, q 2 D 3 .2; 1; 2/, q 3 D 1 .2; 2 ; 1/ (b) The nullspace 3 T T 1T of A contains q 3 (c) b D .A A/ A .1; 2; 7/ D .1; 2/. x 16 The projection p D .aT b=aT a/a D 14a=49 D 2a=7 is closest to b; q 1 D a=kak D a=7 is .4; 5; 2; 2/=7. B D b p D . 1; 4; 4 ; 4/=7 has kB k D 1 so q 2 D B . p 17 p D .aT bp T a/a D .3; 3; 3/ and e D . 2 ; 0; 2/. q 1 D .1; 1; 1/= 3 and q 2 D =a . 1; 0; 1/= 2. 1 18 A D a D .1; 1; 0; 0/I B D b p D . 2 ; 1 ; 1; 0/I C D c pA p B D . 1 ; 1 ; 1 ; 1/. 2 333 Notice the pattern in those orthogonal A ; B ; C . In R5 , D would be . 1 ; 1 ; 1 ; 1 ; 1/. 4444 15 (a) q 1 D 19 If A D QR then AT A D RT QT QR D RT R D lower triangular times upper triangular (this Cholesky factorization of AT A uses the same R as Gram-Schmidt!). The example " # " #  11 1 2 33 1 2 1 D3 2 1 has A D D QR and the same R appears in 03 2 4  2 2    99 30 33 AT A D D D RT R. 9 18 33 03 2 2 (b) True. Qx D x1 q 1 C x2 q 2 . kQx k2 D x1 C x2 because q 1  q 2 D 0. p 21 The orthonormal vectors are q 1 D .1; 1; 1; 1/=2 and q 2 D . 5; 1; 1; 5/= 52. Then b D . 4 ; 3; 3; 0/ projects to p D . 7; 3; 1; 3/=2. And b p D . 1; 3; 7; 3/=2 is orthogonal to both q 1 and q 2 . 22 A D .1; 1; 2/, B D .1; 1; 0/, C D . 1; 1; 1/. These are not yet unit vectors. 20 (a) True Solutions to Exercises 50 23 24 25 26 27 "# "# "# " #" # 1 0 0 100 124 036D You can see why q 1 D 0 , q 2 D 0 , q 3 D 1 . A D 0 0 1 0 1 0 010 005 QR. (a) One basis for the subspace S of solutions to x1 C x2 C x3 x4 D 0 is v1 D .1; 1; 0; 0/, v2 D .1; 0; 1; 0/, v3 D .1; 0; 0; 1/ (b) Since S contains...
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This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.

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