TITLE: Matrix Inverses and Eigenvalues
QUESTION [25 upvotes]: I was working on this problem here below, but seem to not know a precise or clean way to show the proof to this question below. I had about a few ways of doing it, but the statements/operations were pretty loosely used. The problem is as follows:
Show that ${\bf A}^{-1}$ exists if and only if the eigenvalues $ \lambda _i$ ,  $1 \leq i \leq n$ of $\bf{A}$ are all non-zero, and then ${\bf A}^{-1}$ has the eigenvalues given by $ \frac{1}{\lambda _i}$, $1 \leq i \leq n$.
Thanks.

REPLY [2 votes]: Here is a different approach that gives more information.  It shows that the algebraic multiplicities match up as well.  I'll defer to previous answers for the fact that a matrix is invertible iff $0$ is not an eigenvalue, which will be used implicitly below.  I just focus on the eigenvalues of the inverse.
Suppose $A$ is an invertible $n \times n$ matrix.  Assuming (WLOG) we're working in an algebraically closed field, the eigenvalues are the roots of the polynomial $\det(A-\lambda I) = \prod_{1 \leq i \leq n} (\lambda_i - \lambda)$.  Using the properties of determinants, we get:
$$ \det(A^{-1} - \lambda I) = \det(A^{-1}) \det(I - \lambda A) = \det(A^{-1}) ( -\lambda)^n \det(A - \lambda^{-1}I)$$
$$ = \det(A^{-1}) ( -\lambda)^n \prod_{1 \leq i \leq n} (\lambda_i - \lambda^{-1}) = \det(A^{-1}) \prod_{1 \leq i \leq n} (1 - \lambda \lambda_i) $$
$$ = \det(A)^{-1} \prod_{1 \leq i \leq n} \lambda_i \prod_{1 \leq i \leq n} (\lambda_i^{-1} - \lambda) = \det(A)^{-1} \det(A) \prod_{1 \leq i \leq n} (\lambda_i^{-1} - \lambda)$$
$$ = \prod_{1 \leq i \leq n} (\lambda_i^{-1} - \lambda).$$