Eigenvalue and Diagonalization Proofs

Question

I'm having trouble figurintg out how to start the following two proofs

1. If A is invertible and $ \lambda $ is an eigenvalue of A associated to a vector x, then 1/$\lambda$ is an eigenvalue of $A^{-1}$ associated to the same vector x

2. If A is invertible and diagonalizable, then $A^{-1}$ is also diagonalizable

I know a matrix is invertible if it has a non-zero determinant and diagonalizable if the eigenvectors are linearly independent. But I'm not sure how to express these in a general form for any A that satisfies these properties. I don't have that much experience with proofs involving matrices and I'm not sure how to make the matrix generalized for these properties.

Answer 1

Just a matter of applying the relevant formula...

If $Ax=\lambda x$ then $x = \lambda A^{-1} x$ and so $A^{1-} x = { 1\over \lambda} x$.

If $A = V \Delta V^{-1}$, then $A^{-1} = V \Delta^{-1} V^{-1}$.

Answer 2

For (1), $$ A x= \lambda x \rightarrow \frac{1}{\lambda}x=A^{-1}x$$