Find the Eigenvalues of an Operator on Matrices Suppose A is Non Singular and  $\lambda_1,.......,\lambda_n$ are roots of the characteristic polynomial of A
Note : These are Operators on the Vector Space of square n by n Matrices.
$L : X \rightarrow AXA^T$
$S : X \rightarrow AXA^{-1}$
Show that L has Eigenvalues $\lambda_i *\lambda_j$ where $i,j = 1,...n$
Show that S has Eigenvalues $\lambda_i \; / \; \lambda_j$ where $ i,j = 1,...n$
 A: Take $v_i$ an eigenvector associated with the eigenvalue $\lambda_i$. For $V_{i,j}=v_iv_j^T$ (this is an $nxn$ matrix. One has
$$L(V_{i,j})=Av_iv_j^TA^T=(Av_i)(Av_j)^T=\lambda_i\lambda_j(v_iv_j^T)$$
When $A$ is invertible $\lambda_i\neq 0$ and is an eigenvalue of $A^T$ take $w_i$ a corresponding eigenvector. Then $(A^{-1})^{-1}w_i=\frac{1}{\lambda_i}w_i$. Let's compute for $W_{i,j}$
$$S(W_{i,j})=Av_iw_j^TA^{-1}=(Av_i)((A^T)^{-1}w_j)^T=\frac{\lambda_i}{\lambda_j}(v_iw_j^T)$$
A: @ Exc , more generally, consider the function $f:X\in M_n(\mathbb{C})\rightarrow AXB^T$ where the spectra of $A,B$ are $(\lambda_i)_i,(\mu_i)_i$. If we stack a square complex matrix row by row, then $f=A\otimes B$ is a $n^2\times n^2$ matrix.
cf. http://en.wikipedia.org/wiki/Kronecker_product
The key is: if $P,Q$ are invertible, then $A\otimes B$ and $(PAP^{-1})\otimes (QBQ^{-1})$ are similar. Thus we may assume that $A,B$ are upper triangular ; then $A\otimes B$ is upper triangular and its diagonal is composed of the $n^2$ products $\lambda_i\mu_j$.
Example: $\begin{pmatrix}a&b\\0&c\end{pmatrix}\otimes \begin{pmatrix}d&e\\0&f\end{pmatrix}=\begin{pmatrix}ad&ae&*&*\\0&af&*&*\\0&0&cd&ce\\0&0&0&cf\end{pmatrix}$ 
PS: To calculate the spectrum of $g:X\rightarrow AX^TB$ is more complicated. (Consider $g\circ g$, ...)
