Diagonalizing a matrix?

Question

let matrix A=[ 1 1 ; 4 1]. (A 2x2 matrix with (1 1) in first row and (4 1) in the 2nd).
Diagonalize A by finding P and D such that A=PDP^-1.

Hence compute A^50.

Is it true that A^-1=(P)(D^-1)(P^-1)

please try to send me the final answers as well as the workings... thanks :)

Answer 1

a) The matrix P is the matrix whose columns are the eigenvectors of A

b) To find A^50, think about writing PDP^-1 times itself 50 times, and notice how all the P's and P-inverses cancel. Raising D to a power is easy...just raise the numbers on the diagonal to that power.

c) yes, your identity is true. to prove it, use the theorem that says (ST)^-1 = (T^-1)(S^-1) , which of course generalizes to 3 matrices, or any finite number of matrices.

Answer 2

The diagonal entries of D are the eigenvalues of A, and the columns of P are the eigenvectors. It is important to note that not all matrices are diagonalisable if the eigenvalues are repeated.

The eigenvalues of A are 3 and -1 (found by setting det(A-lambda I) = 0); corresponding eigenvectors are [1 2] and [1 -2]. This A is diagonalisable because the eigenvalues are distinct.

So D = diag[3, -1] and P = [[1 2]^t, [1 -2]^t]. This makes P^-1 equal to [[1/2 1/2]^t [1/4 -1/4]^t], and the factorisation of A is

A = PDP^-1
= |1 1 | | 3 0 | | 1/2 1/4 |
|2 -2 | | 0 -1 | | 1/2 -1/4 |