11.4. EXERCISES WITH ANSWERS
197
2 0 −1 0 . Find a formula for An where n is an integer. −2 1 1 2 0 First you find the eigenvectors and eigenvalues. 0 −1 0 , eigenvectors: 0 −2 1 0 −1 1 0 , 0 ↔ 1, 1 ↔ −1. 1 1 0
1 6. Here is a matrix. A = 0 0
The matrix, S used to diagonalize the matrix is the columns of S. Then S −1 is given by 1 1 S −1 = 0 −1 0 1 Then S −1 AS equals
obtained by letting these vectors be 0 1 0
1 0 1 2 0 1 0 −1 1 0 −1 0 0 0 1 0 0 −2 1 0 1 1 0 0 = 0 1 0 ≡D 0 0 −1
1 0 0
−1 1 1
Then A = SDS −1 and An = SDn S −1 . Now it is easy to find Dn . 1 0 0 0 Dn = 0 1 n 0 0 (−1) Therefore, An
=
=
1 0 −1 0 0 1 0 1 1 n 1 1 − (−1) n 0 (−1) n 0 −1 + (−1)
7. Supposethe eigenvalues of eλ1 t · · · 0 .. .. . At .. e =S . . 0
···
1 0 0 1 0 0 1 0 n 0 0 (−1) 0 0 0 . 1
1 0 −1 1 1 0
A are λ1 , · · · , λn and that A is nondefective. Show that −1 S where S is the matrix which satisfies S −1 AS = D.
eλ n t
The diagonal matrix, D has the same characteristic equation as A why? and so it has the same eigenvalues. However the eigenvalues of D are the diagonal entries and so the diagonal entries of D are the eigenvalues of A. Now S −1 tAS = tD