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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!

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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
    
The well know «famous among the people who know about it» paper :) Thanks for the reference! –  Mariano Suárez-Alvarez Jan 4 '12 at 6:01
    
@Mariano: Well, I specified the paper reasonably tightly, but it's in Mathscinet as MR0072872. –  Geoff Robinson Jan 4 '12 at 15:51
    
I was not complaining about that or anything, really :) –  Mariano Suárez-Alvarez Jan 4 '12 at 16:18
    
@Mariano: Sure, I know, but I thought I might as well put in a bit more about it. –  Geoff Robinson Jan 4 '12 at 17:11
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