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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.

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    $\begingroup$ 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. $\endgroup$ Dec 4, 2014 at 4:37
  • $\begingroup$ 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. $\endgroup$ Dec 4, 2014 at 7:59
  • $\begingroup$ 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. $\endgroup$
    – Tian An
    Dec 4, 2014 at 15:10
  • $\begingroup$ I mean, there has to be one by symmetric function theory, but it's not very nice. What do you want to do? $\endgroup$ Dec 5, 2014 at 5:04
  • $\begingroup$ 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 $p_{A\otimes B}$. $\endgroup$
    – Tian An
    Dec 8, 2014 at 18:39

2 Answers 2

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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 $x^4+2x^3-479x^2+420x+44100=(x-15)(x-14)(x+10)(x+21)$.

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  • $\begingroup$ If I understand correctly, the resultant is a polynomial of the coefficients, but not in the original variable. How can I recover the polynomial with roots the products of the eigenvalues with it? Thanks! $\endgroup$
    – Tian An
    Dec 8, 2014 at 20:44
  • $\begingroup$ I change variables before calculating the resultant, so that the resultant itself ends up as a polynomial in the original variable -- I added an example above. $\endgroup$
    – ahulpke
    Dec 9, 2014 at 2:02
  • $\begingroup$ Ah. I get now; it's exactly what I was looking for. Thanks very much! $\endgroup$
    – Tian An
    Dec 9, 2014 at 4:00
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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 $(\det B)^n \prod_{j=1}^m P_A(x/\mu_j)$.

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