Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# Has determinant 1 so the house is ipped and sheared

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Unformatted text preview: hat diagonal matrix AT A are the columns of I , so V D I in the SVD. Then the ui are Avi =i which is the unit vector wi =i . The SVD of this A with orthogonal columns is A D U †V T D .A† 1 /.†/.I /: 2 2 12 Since AT D A we have 1 D 2 and 2 D 2 . But 2 is negative, so 1 D 3 and 1 2 13 14 15 16 17 2 D 2. The unit eigenvectors of A are the same u1 D v1 as for AT A D AAT and u2 D v2 (notice the sign change because 2 D 2 , as in Problem 4). Suppose the SVD of R is R D U †V T . Then multiply by Q to get A D QR. So the SVD of this A is .QU /†V T . (Orthogonal Q times orthogonal U D orthogonal QU .) The smallest change in A is to set its smallest singular value 2 to zero. See # 7. The singular values of A C I are not j C 1. They come from eigenvalues of .A C I /T .A C I /. This simulates the random walk used by Google on billions of sites to solve Ap D p. It is like the power method of Section 9:3 except that it follows the links in one “walk” where the vector pk D Ak p0 averages over all walks. p p A D U †V T D Œcosines including u4  diag.sqrt.2 2; 2; 2 C 2// Œsine matrixT . AV D U † says that differences of sines in V are cosines in U times  ’s. The SVD of the derivative on Œ0;  with f .0/ D 0 has u D sin nx ,  D n, v D cos nx ! Problem Set 7.1, page 380 1 With w D 0 linearity gives T .v C 0/ D T .v/ C T .0/. Thus T .0/ D 0. With c D 1 linearity gives T . 0/ D T .0/. This is a second proof that T .0/ D 0. 2 Combining T .c v/ D cT .v/ and T .d w/ D d T .w/ with addition gives T .c v C d w/ D cT .v/ C d T .w/. Then one more addition gives cT .v/ C d T .w/ C eT .u/. 3 (d) is not linear. Solutions to Exercises 76 4 (a) S.T .v// D v (b) S.T .v1 / C T .v2 // D S.T .v 1 // C S.T .v2 //. 5 Choose v D .1; 1/ and w D . 1; 0/. Then T .v/ C T .w/ D .v C w/ but T .v C w/ D .0; 0/. 6 (a) T .v/ D v=kvk does not satisfy T .v C w/ D T .v/ C T .w/ or T .c v/ D cT .v/ (d) satisﬁes T .c v/ D cT .v/. (b) and (c) are linear 7 (a) T .T .v // D v (b) T .T .v// D v C .2; 2/ (c) T .T .v // D v (d) T .T .v// D T .v/. 8 (a) The range of T .v1 ; v2 / D .v1 v2 ; 0/ is the line of vectors .c; 0/. The nullspace is the line of vectors .c; c/. (b) T .v1 ; v2 ; v3 / D .v1 ; v2 / has Range R2 , kernel {(0; 0; v3 )} (c) T .v/ D 0 has Range f0g, kernel R2 (d) T .v1 ; v2 / D .v1 ; v1 / has Range = multiples of .1; 1/, kernel = multiples of .1; 1/. 9 If T .v1 ; v2 ; v3 / D .v2 ; v3 ; v1 / then T .T .v// D .v3 ; v1 ; v2 /; T 3 .v/ D v; T 100 .v/ D T .v /. 10 (a) T .1; 0/ D 0 (b) .0; 0; 1/ is not in the range n m (c) T .0; 1/ D 0. 11 For multiplication T .v/ D Av: V D R , W D R ; the outputs ﬁll the column space; v is in the kernel if Av D 0. 12 T .v/ D .4; 4/I .2; 2/I .2; 2/; if v D .a; b/ D b.1; 1/C a 2 b .2; 0/ then T .v/ D b.2; 2/C .0; 0/. 13 The distributive law (page 69) gives A.M1 C M2 / D AM1 C AM2 . The distributive law over c ’s gives A.cM / D c.AM /. 14 This A is invertible. Multiply AM D 0 and AM D B by A 1 to get M D 0 and B . The kernel contains only the zero matrix M D 0....
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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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