# Matrix exp. question

1. Nov 14, 2006

### scienceman2k9

I need to find e^A of this 2x2 matrix....A= 1 1
-1 -1

When I do det(A-(lambda)I) I get 0 for the eigenvalues, which makes no sense. Am I doing something wrong?

2. Nov 14, 2006

### NateTG

Probably not. If the matrix is singular then at least one of the eigenvalues will be zero
Hmm...
$$det(A-I \lambda )=(-\lambda+1)(-\lambda-1)+1=\lambda^2$$
So $0$ is a double root, and both of the eigenvalues are zero.

Why don't you see what happens if you apply this matrix to a vector?

Last edited: Nov 14, 2006