Introduction to Linear algebra-Strang-Solutions-Manual_ver13

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 forthe 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...
View Full Document

## 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.

Ask a homework question - tutors are online