Linear transformation Eigenvalues for matrices: M to AMB Determine the trace and determinant of the linear operator (on the space $\mathbb{F^{n\times n}}$) that sends the matrix $M\to AMB$ where $A$ and $B$ are $n\times n$ matricies 
I know this question has been asked before. But I am interested in using the eigenvalues to solve this problem. What could be the eigenvalues/eigenvectors of this problem. I know how to use brute force for $2$ by $2$ matrixes. But stuck on making a geberelazitions for eigenvalues.
 A: There is a useful identity that allows us to vectorize the matrix $M$ and interpret it as a vector in $\mathbb{F}^{n\times n}$.  This is given by
$$
\text{vec}(AMB) \;\; =\;\; (B^T\otimes A)\text{vec}(M)
$$
where $\text{vec}(X)$ is $n^2\times 1$ and is constructed by concatenating the columns of $X$ where $X$ is $n\times n$, and $\otimes$ is the Kronecker product given by:
$$
X\otimes Y \;\; =\;\; \left [ \begin{array}{cccc}
y_{11}X & y_{12}X &\ldots  & y_{1n}X \\
y_{21}X & y_{22}X &\ldots  & y_{2n}X \\
\vdots & \vdots & \ddots & \vdots \\
y_{m1}X & y_{n2}X &\ldots  & y_{mn}X \\
\end{array}\right ]
$$
where $X$ and $Y$ can be matrices of arbitrary size.  You can then compute the trace and determinant of $B^T\otimes A$ which will be 
$$
\text{Tr}(B^T\otimes A) \;\; =\;\; \text{Tr}(A)\text{Tr}(B) \hspace{3pc} \det(B^T\otimes A) \;\; =\;\; \det(B)^n\det(A)^n.
$$
A: Consider the case where $A$ and $B$ each have distinct eigenvalues. This case
is generic enough. Let $\alpha_1,\ldots,\alpha_n$ be the eigenvalues
of $A$ and $\beta_1,\ldots,\beta_n$ be those of $B$. If we have column eigenvectors
$v_1,\ldots,v_n$ of $A$ (so $Av_i=\alpha_iv_i$) and row eigenvectors
$w_1,\ldots,w_n$ of $A$ (so $w_jV=\beta_j w_j$) then the matrices
$v_iw_j$ are eigenvectors of your transformation with eigenvalues $\alpha_i\beta_j$. Therefore its trace is
$$\sum_{i,j=1}^n\alpha_i\beta_j=\text{tr}(A)\text{tr}(B)$$
and its determinant is
$$\prod_{i,j=1}^n\alpha_i\beta_j=\det(A)^n\det(B)^n.$$
