This was asked before on [stackexchange](https://math.stackexchange.com/questions/666122/kronecker-product-and-characteristic-polynomial) but no answer was given. The question is the following:

Let $A$ and $B$ be matrices in $GL(n)$ and $GL(m)$ respectively. Their tensor product $A\otimes B$ is defined explicitly by the Kronecker product:
$$A\otimes B=\begin{bmatrix}a\_{11}B&\dots&a\_{1n}B\\\vdots&\ddots&\vdots\\a\_{n1}B&\dots&a\_{nn}B\end{bmatrix}\in GL(nm).$$
Question: is there a known expression of its characteristic polynomial in terms of those of $A$ and $B$? It seems there might not be one, but I would like to be proven wrong.

(In contrast, the direct sum $A\oplus B=\begin{bmatrix}A&0\\0&B\end{bmatrix}$ has characteristic polynomial $p\_A\cdot p\_B.$)

EDIT: I've worked out an example for $n=m=2$: Given $2\times 2$ matrices $A$ and $B$ with characteristic polynomial $p\_A=t^2-a\_1t+a\_2$ and $p\_B=t^2-b\_1t+b\_2$, I find the coefficients of $p\_{A\otimes B}=t^4-c\_1t^3+c\_2t^2-c\_3t+c\_4$ can be expressed as:
\begin{align}
c\_1&=a\_1b\_1\\
c\_2&=a\_1^2b\_2+b\_1^2a\_2-2a\_2b\_2\\
c\_3&=b\_1b\_2a\_1a\_2\\
c\_4&=a\_2^2b\_2^2
\end{align}
Similarly for $n=2,m=3$,
\begin{align}
c\_1&=a\_1b\_1\\
c\_2&=a\_1^2b\_2+b\_1^2a\_2-2a\_2b\_2\\
c\_3&=b\_1b\_2(a\_1a\_2-3a\_3)+a\_3b\_1^3\\
c\_4&=a\_1a\_3(b\_1b\_2-2b\_2^2)+a\_2^2b\_2^2\\
c\_5&=b\_1b\_2^2a\_1a\_2\\
c\_6&=a\_2^3b\_2^2
\end{align}
It is clear that in general $c\_1=a\_1b\_1$ and $c\_n=a\_n^{\text{deg}(p\_B)}b\_n^{\text{deg}(p\_A)}$, $c\_2$ also seems to be consistent, but a general formula for $c\_i$ eludes me still.