Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# 1 has d 0 0 and only one independent eigenvector x d i

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Unformatted text preview: C 6 A4 t 3 C    D A.I C At C 2 A2 t 2 C 6 A3 t 3 C    /. This is exactly Ae At , the derivative we expect.  1 Bt 2 19 e D I C Bt (short series with B D 0) D 0   4t 0 . Derivative D 1 0  4 D B. 0 20 The solution at time t C T is also e A.t CT / u.0/. Thus e At times e AT equals e A.t CT / . 21  1 0  4 1 D 0 0 4 1  10 00  1 0  4 1 ; 1 0 4 1  et 0 0 1  22 A2 D A gives e At D I C At C 1 At 2 C 1 At 3 C    D I C .e t 2 6  t  4 e 4e t 4 D . 1 0 1 t  e et 1 1/A D . 0 1    e 4.e 1/ 1 4 B 23 e D from 21 and e D from 19. By direct multiplication 0 1 0 1   e0 e A e B ¤ e B e A ¤ e ACB D . 01       t 1 3t  1 .e et / e 11 11 10 1 2 . Then e At D 2 . 24 A D D 1 03 02 03 0 0 e 3t 2  2   13 13 2 25 The matrix has A D D D A. Then all An D A. So e At D 00 00 t  e 3.e t 1/ 2 t I C .t C t =2Š C    /A D I C .e 1/A D as in Problem 22. 0 0 A  1 0 26 (a) The inverse of e At is e At At To see e x , write .I C At C (b) If Ax D x then e At x D e t x and e t ¤ 0. C    /x D .1 C t C 1 2 t 2 C    /x D e t x . 2 1 22 At 2 t 27 .x; y/ D 4t ; e 4/ is a growing solution. The correct matrix for the exchanged u D .e 2 . It does have the same eigenvalues as the original matrix. 0     1 t 11 28 Centering produces U nC1 D U n . At t D 1, has  D 10 t 1 .t /2 e i=3 and e i =3 . Both eigenvalues have 6 D 1 so A6 D I . Therefore U 6 D A6 U 0 comes exactly back to U 0 .   1 2n 2n First A has  D ˙i and A4 D I . n 29 A D . 1 /n Linear growth. 2n 2n C 1 Second A has  D 1; 1 and   1 1 a2 2a U n. 30 With a D t =2 the trapezoidal step is U nC1 D 2 a 1 a2 1 C a2 .y; x/ is 2 4 That matrix has orthonormal columns ) orthogonal matrix ) kU nC1 k D kU n k 31 (a) .cos A/x D .cos /x (b) .A/ D 2 and 0 so cos  D 1; 1 and cos A D I (c) u.t / D 3.cos 2 t /.1; 1/C1.cos 0t /.1; 1/ Œ u 0 D Au has exp; u 00 D Au has cos  Solutions to Exercises 67 Problem Set 6.4, page 337 Note A way to complete the proof at the end of page 334, (perturbing the matrix to produce distinct eigenvalues) is now on the course website: “Proofs of the Spectral Theorem.” math.mit.edu/linearalgebra. &quot; #&quot; # 136 0 1 2 D 1 .A C AT / C 1 .A AT / 2 2 0 3 1 AD 3 3 3 C 1 D symmetric C skew-symmetric: 635 2 3 0 2 .AT CA/T D AT C T .AT /T D AT CA. When A is 6 by 3, C will be 6 by 6 and the triple product AT CA is 3 by 3. p p p 3  D 0; 4; 2; unit vectors ˙.0; 1; 1/= 2 and ˙.2; 1; 1/= 6 and ˙.1; 1; 1/= 3.     10 0 1 2 5 in ƒ D ,xD and have to be normalized to unit 0 5 2 1   11 2 vectors in Q D p . 2 1 5 &quot; # 2 1 2 1 The columns of Q are unit eigenvectors of A 2 2 1. QD Each unit eigenvector could be multiplied by 1 3 1 2 2   9 12 AD has  D 0 and 25 so the columns of Q are the two eigenvectors: 12 16   :8 :6 QD or we can exchange columns or reverse the signs of any column. :6 :8   12 (a) has  D 1 and 3 (b) The pivots have the same signs as the ’s (c) trace 21 D 1 C 2 D 2, so A can’t have two negative eigenvalues.  ...
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