Eigenvalues of Invertible Matrix Suppose I have two square matrices $A, B$ with the same size $n\times n,$ and $AB = I = BA.$ What conclusions could we draw for the eigenvalues and eigenvectors of matrix $A?$
I only know that $A$ is an invertible matrix, and it has non-zero eigenvalues. I don't think $A$ can have $n$ distinct eigenvalues to say that it is diagonalizable.
 A: Suppose $A$ is an invertible square matrix and $B$ is its inverse.
Eigenvalues of $B$ are inverse to those of $A$ (and vice versa) and $A,B$ share eigenvectors corresponding to the pairs of inverse eigenvalues $(\lambda,\lambda^{-1}).$
To prove it,
let $\lambda\neq 0$ be an eigenvalue of $A$ and $u$ a right eigenvector corresponding to $\lambda.$ This writes
$$u=(BA)u=B(Au)=B(\lambda u)=\lambda Bu.$$ Therefore $$Bu={\lambda}^{-1}u.$$
Similarly, if $v$ is a left eigenvector corresponding to the eigenvalue $\lambda$ of $A,$ then $v$ is also a left eigenvector of $B$ corresponding to $\lambda^{-1}.$
This proves also that inverse matrices have equal algebraic multiplicity and equal geometric multiplicity.
EDIT
Left (or "row") eigenvectors corresponding to the eigenvalue $\mu$ satisfy the relation $$v^TA=\mu v^T.$$  Let us prove that the eigenvalues corresponding to the left or right eigenvectors are the same. (So, the name "eigenvalue" without specifying the side is legitimate.)
For, realize that $A$ and its transpose $A^T$ share eigenvalues because they have the same charasteristic polynomial.
Let $(\lambda,u)$ be an eigenpair of $A.$
$$\begin{aligned}Au=\lambda u \iff &(Au)^T=(\lambda u)^T\\ \iff & u^T A^T=\lambda u^T \end{aligned}$$
Therefore, there exists a row vector $v^T$ such that  $$v^T A=\lambda v^T.$$
The rest of the proof can be completed analogously.
