# How to find the Jordan canonical form of tensor products.

Let $k$ be a field, $A \in M_{m\times m}(k)$ be a single Jordan block with eigenvalue $a$, and $B \in M_{n\times n}(k)$ be a single Jordan block with eigenvalue $b$. $A$ and $B$ together define a linear transformation $$A\otimes B : k^{m\times n} \to k^{m\times n}$$ and my question is

what is the Jordan canonical form of $A\otimes B$? (The Jordan canonical form of $A\otimes B$ exists because the characteristic polynomial of $A\otimes B$ splits and equals $(t-ab)^{mn}$)

One can tickle this question directly by finding the $A\otimes B$ invariant subspaces of $k^{m\times n}$:

Assume $m\geq n$, let $\{e_i\}$ be a basis of $k^m$, and $\{f_j\}$ be a basis of $k^n$, then $\{e_i\otimes f_j\}$ form a basis of $k^{m\times n}$. In the case $a=b=0$, $k^{m\times n}$ decomposes into the direct sum $$\bigoplus_{l=n-m}^{m-n} V_l$$ where $V_l=\mathrm{span}\{e_i\otimes e_j|i-j=l\}$, and $A\otimes B$ restricted to each $V_l$ is a single Jordan block. The case exactly one of $a,b$ equals zero is not much harder. But the case $ab\neq 0$ is much more difficult.

Another way to do find the elementary divisors of the map $$k[t]^{m\times n}\xrightarrow{tI-A\otimes B} k[t]^{m\times n}$$ using the formula $$d_i=\gcd(\{i\times i\text{ minor of }A\otimes B\})$$ but I can't find a good way to calculate the $\gcd$'s.

Any help or hints would be appreciated. Thank you!

## 1 Answer

This is only a couple of hints/suggestions, but a bit long for the comments boxes.

First note that for any linear operator $\alpha$ on a finite dimensional $\mathbb{k}$-vector space $V$ and scalar $c\not=0$ we have that $c\alpha$ has, for each $\lambda\in\mathbb{k}$ and each $r\in\mathbb{N}$, the same number of $c\lambda$-$(r\times r)$ Jordan blocks as $\alpha$ has $\lambda$-$(r\times r)$ Jordan blocks. This is because the sequence of ranks $\{\rho\left((\alpha-\lambda)^s\right)\}_{s=0}^{\dim V}$ determines the pattern of $\lambda$-$(r\times r)$ blocks.

By distributing $ab\not=0$ across the tensor product we can assume without loss of generality that $a=b=1$.

Secondly, by the above remark about ranks, all that is needed now (!) is to compute the ranks of the powers of $(I_m+J_1)\otimes(I_n+ J_2)-I_m\otimes I_n$. Suitably ordering the basis, this is a block matrix with $J_2$s on the main diagonal and $I+J_2$ on the superdiagonal, and zeros elsewhere. I've only done very small cases to see what happens as we take powers, but it doesn't look infeasible: especially as Jordan Block of Kronecker Product. tells us what answer we're looking for.

I repeat this is not offered as a solution, just an alternative strategy.