Introduction to Linear algebra-Strang-Solutions-Manual_ver13

W 1 1 a1 6 ad has real eigenvalues a c 1 and a 1

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Unformatted text preview: 10 11 ing 2  1 D 0 as the Cayley-Hamilton Theorem predicts. When A D SƒS 1 is diagonalizable, the matrix A j I D S.ƒ j I /S 1 will have 0 in the j; j diagonal entry of ƒ j I . In the product p.A/ D .A 1 I /    .A n I /, each inside S 1 cancels S . This leaves S times (product of diagonal matrices ƒ j I ) times S 1 . That product is the zero matrix because the factors produce a zero in each diagonal position. Then p.A/ D zero matrix, which is the Cayley-Hamilton Theorem. (If A is not diagonalizable, one proof is to take a sequence of diagonalizable matrices approaching A.) Comment I have also seen this reasoning but I am not convinced: Apply the formula AC T D .det A/I from Section 5.3 to A I with variable . Its cofactor matrix C will be a polynomial in , since cofactors are determinants: .A I / cof .A I /T D det.A I /I D p./I: “For fixed A, this is an identity between two matrix polynomials.” Set  D A to find the zero matrix on the left, so p.A/ D zero matrix on the right—which is the CayleyHamilton Theorem. I am not certain about the key step of substituting a matrix for . If other matrices B are substituted, does the identity remain true? If AB ¤ BA, even the order of multiplication seems unclear : : : 33  D 2; 1; 0 are in ƒ and the eigenvectors are in S (below). Ak D S ƒk S " 2 1 1 1 1 1 # 0 1 1 ƒk 6 1 " 2 2 0 1 2 3 # 1 2k 2D 6 3 " 4 2 2 # 22 . 1 /k 11C 3 11 " 1 1 1 1 1 1 1 is 1 1 1 # Check k D 4. The .2; 2/ entry of A4 is 24 =6 C . 1/4 =3 D 18=6 D 3. The 4-step paths that begin and end at node 2 are 2 to 1 to 1 to 1 to 2, 2 to 1 to 2 to 1 to 2, and 2 to 1 to 3 to 1 to 2. Much harder to find the eleven 4-step paths that start and end at node 1. Solutions to Exercises 64 34 If AB D BA, then B has the same eigenvectors .1; 0/ and .0; 1/ as A. So B is also diagonal b D  D 0. The nullspace forthe following equation is 2-dimensional: c         10 ab ab 10 0 b 00 AB BA D D D . The 02 cd cd 02 c 0 00 coefficient matrix has rank 4 2 D 2. p 35 B has  D i and i , so B 4 has 4 D 1 and 1 and B 4 D I . C has  D .1 ˙ 3i /=2. This is exp.˙ i=3/ so 3 D 1 and 1. Then C 3 D I and C 1024 D C .   cos  sin  36 The eigenvalues of A D are  D e i and e i  (trace 2 cos  and sin  cos  det D 1). Their eigenvectors are .1; i / and .1; i /:     1 1 e i n i 1 An D Sƒn S 1 D =2i ii i 1 e i n  i n   .e C e i n /=2    cos n sin n D D : sin n cos n .e i n e i n /=2i    Geometrically, n rotations by  give one rotation by n . 37 Columns of S times rows of ƒS 1 will give r rank-1 matrices .r D rank of A/. 38 Note that ones.n/  ones.n/ D n  ones.n/. This leads to C D 1=.n C 1/. AA 1 D .eye.n/ C ones.n//  .eye.n/ C C  ones.n// D eye.n/ C .1 C C C C n/  ones.n/ D eye.n/: Problem Set 6.3, page 325 1 u1 D e 4t     1 1 1 t 4t 1 t , u2 D e . If u.0/ D .5; 2/, then u.t / D 3e C 2e . 0 1 0 1 2 z.t / D 2e t ; then dy=dt D 4y Problem 1. 6e t with y.0/ D 5 gives y.t / D 3e 4t C 2e...
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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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