# Characteristic polynomial of Kronecker/tensor product

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.

• Since the set of eigenvalues of $A \otimes B$ is just the set of products (with multiplicities) of eigenvalues of $A$ with those of $B$, the characteristic polynomial is calculable if the eigenvalues of $A$ and of $B$ respectively. But this doesn't really help if you want an explicit expression (suggesting that there is nothing simple). There are also corresponding (horrible) formulas for the symmetric functions. Dec 4, 2014 at 4:37
• The case of direct sums is nice because the exterior power functors play nicely with direct sums; in fact $\Lambda^{\bullet}(-)$ converts direct sums to (graded) tensor products. Their behavior with respect to tensor products is substantially more complicated. Dec 4, 2014 at 7:59
• Abstractly, I understand that we get it from the product of the eigenvalues of $A$ and $B$; but unfortunately I'm looking to see if there is an explicit expression. Dec 4, 2014 at 15:10
• I mean, there has to be one by symmetric function theory, but it's not very nice. What do you want to do? Dec 5, 2014 at 5:04
• I would prefer a closed formula for the coefficients, but at the very least a statement that given the coefficients (not eigenvalues) of $p_A$ and $p_B$ I can construct the coefficients of $p_{A\otimes B}$. Dec 8, 2014 at 18:39

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 $x^4+2x^3-479x^2+420x+44100=(x-15)(x-14)(x+10)(x+21)$.
If the characteristic polynomials of $A$ and $B$ factor as $P_A(x) = \prod_{i=1}^n (x - \lambda_i)$ and $P_B(x) = \prod_{j=1}^m (x - \mu_j)$, then the characteristic polynomial of $A \otimes B$ is $$P_{A \otimes B}(x) = \prod_{i=1}^n \prod_{j=1}^m (x - \lambda_i \mu_j)$$ If all $\lambda_i \ne 0$ this could be written as $$\prod_{i=1}^n \lambda_i^m \prod_{j=1}^m (x/\lambda_i - \mu_j) = (\det A)^m \prod_{i=1}^n P_B(x/\lambda_i)$$ or similarly if all $\mu_i \ne 0$ as $(\det B)^n \prod_{j=1}^m P_A(x/\mu_j)$.