Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# C 0 d 05 0c0d 1000 38 the matrix for t in this basis

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Unformatted text preview: nput basis. Solutions to Exercises 78 4 The third derivative matrix has 6 in the .1; 4/ position; since the third derivative of x 3 is 6. This matrix also comes from AB . The fourth derivative of a cubic is zero, and B 2 is the zero matrix. 5 T .v1 C v2 C v3 / D 2w1 C w2 C 2w3 ; A times .1; 1; 1/ gives .2; 1; 2/. 6 v D c.v2 v 3 / gives T .v/ D 0; nullspace is .0; c; c /; solutions .1; 0; 0/ C .0; c; c /. 7 .1; 0; 0/ is not in the column space of the matrix A, and w1 is not in the range of the linear transformation T . Key point: Column space of matrix matches range of transformation. 8 We don’t know T .w/ unless the w’s are the same as the v’s. In that case the matrix is A2 . 9 Rank of A D 2 D dimension of the range of T . The outputs Av (column space) match the outputs T .v/ (the range of T ). The “output space” W is like Rm : it contains all outputs but may not be ﬁlled up. " # "# "# 100 1 1 1D 10 The matrix for T is A D 1 1 0 . For the output 0 choose input v D 111 0 0 "# 1 A 1 0 . This means: For the output w1 choose the input v1 v2 . 0 " # 1 00 1 1 1 0 so T 1 .w1 / D v1 v2 ; T 1 .w2 / D v2 v3 ; T 1 .w3 / D v3 . 11 A D 0 11 The columns of A 1 describe T 1 from W back to V . The only solution to T .v/ D 0 is v D 0. 12 (c) T 1 .T .w1 // D w1 is wrong because w1 is not generally in the input space. 13 (a) T .v1 / D v2 ; T .v2 / D v1 is its own inverse (b) T .v1 / D v1 ; T .v2 / D 0 has T 2 D T (c) If T 2 D I for part (a) and T 2 D T for part (b), then T must be I .       1 2 3 1 21 . must be 2A D inverse of (a) (c) A (b) 14 (a) 6 3 5 2 53       0 r s 1 rs to and ; this is the “easy” direcand 15 (a) M D transforms u 0 1 t tu  1 ab tion. (b) N D transforms in the inverse direction, back to the standard cd basis vectors. (c) ad D bc will make the forward matrix singular and the inverse impossible.   1   3 1 10 21 D . 16 M W D 7 3 12 53 17 Recording basis vectors is done by a Permutation matrix. Changing lengths is done by a positive diagonal matrix. 18 .a; b/ D .cos ; sin  /. Minus sign from Q 1 D QT . Solutions to Exercises      11 a 5 19 M D ; D D ﬁrst column of M 45 b 4    11 . 45 79 1  1 D coordinates of in basis 0 1 x 2 ; w3 .x/ D 2 .x 2 x /; y D 4w1 C 5w2 C 6w3 . # " # 0 1 0 1 11 0 :5 : v’s to w’s: inverse matrix D 1 0 0 . The key w’s to v’s: :5 :5 1 :5 1 11 idea: The matrix multiplies the coordinates in the v basis to give the coordinates in the w basis. 2 32 3 2 3 A 4 1 a a2 4 1 b b 2 5 4 B 5 D 4 5 5. This The 3 equations to match 4; 5; 6 at x D a; b; c are 1 c c2 C 6 Vandermonde determinant equals .b a/.c a/.c b /. So a; b; c must be distinct to have det ¤ 0 and one solution A; B; C . The matrix M with these nine entries must be invertible. Start from A D QR. Column 2 is a2 D r12 q 1 C r22 q 2 . This gives a2 as a combination of the q ’s. So the change of basis matrix is R . Start from A D LU . Row 2 of A is `21 (row 1 of U / C `22 (row 2 of U ). The change of basis matr...
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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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