# Centralizer of $M \otimes M$

Let $k$ be a field, $V$ a finite-dimensional $k$-vectorspace and $M \in End(V)$. How can I determine $Z$, the centralizer of $M \otimes M$ in $End(V) \otimes End(V)$?

For example, if $$M=[[1,0],[0,2]],$$ then $M$ is 6-dimensional, consisting of block matrices of shape 1,2,1.

I was confused at first, because this seems to be a contradiction to the fact that the centralizer of a subalgebra of the form $A \otimes B$ is just the tensor product of the centralizers of $A$ and $B$; but here we are considering only the element $M \otimes M$, not $A \otimes A$, where $A$ is the subalgebra generated by $M$.

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This looks like it can get complicated. Generically, at least over an algebraically closed field, a matrix $M$ will have distinct eigenvalues $m_1,\ldots,m_n$, and generically $M\otimes M$ will have distinct eigenvalues $m_1^2,\ldots,m_n^2$ with multiplicity one, and $m_1m_2,m_1m_3,\ldots,m_{n-1}m_n$ with multiplicity two. Thus the centralizer will have dimension $n+4{n\choose 2}=2n^2-n$.
But there are many degenerate cases: for instance if $M$ has eigenvalues $1,a,\ldots,a^{n-1}$ then $M\otimes M$ will have eigenvalues $1,a,\ldots,a^{2n-2}$ with multiplicities $1,2,\ldots n-1,n,n-1,\ldots,1$. Things can get more complicated still.
Then $M$ might have non-trivial Jordan blocks, and then the real fun starts!
It's reasonably well understood (and in the literature in many places, eg in work by B. Srinivasan, circa 1956) how to caclulate the Jordan blocks of $M \otimes M$ from that of $M.$ The famous (at least among group theorists) Hall-Higman paper (PLMS,1956) computes the centralizer of an element a single eigenvalue $1$ but arbitrary Jordan normal form. This suffices to do the general case ( as Robin suggests, the answer is far from pretty). – Geoff Robinson Jan 4 '12 at 5:45