If v is an eigenvector of A with corresponding eigenvalue λ, show that v is an eigenvectors of A^-1. What is its corresponding eigenvalue?
This should be rather easy. The notation, however, depends on whether you write vectors in rows or in columns.
If I use row vectors, then v being eigenvector means
$\displaystyle v.A=\lambda.v$.
What do you get if you multiply this equality by $\displaystyle A^{-1}$ from the right?
(Obviously, you assume that A is invertible, even if you did not write this assumption explicitly.)
$\displaystyle A\mathbf{x}=\lambda\mathbf{x}$
Multiple by A inverse on the left and move lambda left since it is a scalar.
$\displaystyle A^{-1}A\mathbf{x}=\lambda A^{-1}\mathbf{x}$
A inverse A is I.
$\displaystyle \mathbf{x}=\lambda A^{-1}\mathbf{x}$
Divide by lambda.
$\displaystyle \frac{1}{\lambda}\mathbf{x}=A^{-1}\mathbf{x}$