This was asked before on stackexchange 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.
Best Answer
As David Handleman observed, you need (assuming you are over a splitting field) simply the polynomial that has the products of eigenvalues as roots. Using the resultant, you could calculate this polynomial as $\mbox{res}_y(P_A(y),P_B(x/y)\cdot y^m)$. (This is a polynomial in $x$.)
For example, let $P_A(x)=(x-2)(x+3)=x^2+x-6$ and $P_B(x)=(x+5)(x-7)=x^2-2x-35$. Then $P_B(x/y)\cdot y^2=x^2-2xy-35y^2$ and the resultant becomes $x^4+2x^3-479x^2+420x+44100=(x-15)(x-14)(x+10)(x+21)$.