Finding e^A for 2x2 Matrix: Issues with Det(A-(Lambda)I)

[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?

 

[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?