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Let $\mathbb K$ be a field. Let $A\in M_n(\mathbb K)$ be the matrix of a semisimple linear operator (that is, $A$ is diagonalizable in the algebraic closure of $\mathbb K$).

Is it true that the centralizer of $A$ in $M_n(\mathbb K)$ can be decomposed into

$$C_{M_n({\mathbb K})}(A)= M_{n_1}(\mathbb K)\otimes_{\mathbb K}\ldots\otimes_{\mathbb K} M_{n_r}(\mathbb K), $$

where the $n_i$ are the size of the Jordan blocks in the Jordan Canonical Form of $A$ in the algebraic closure of $\mathbb K$?

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Aren't all $n_i$ equal to $1$ if $A$ is diagonalizable? – tomasz Sep 15 '12 at 18:12
@tomasz: Dear tomasz, I think that what is meant is that $n_i$ is the multiplicity of the $i$th eigenvalue of $A$ over the algebraic closure. Regards, – Matt E Sep 15 '12 at 19:16
Well, in any case, it will probably be $\oplus$ and not $\otimes$. This might be worth looking at: – tomasz Sep 15 '12 at 19:55

Yes. In a less coordinate-dependent sense, "diagonalizability" of an operator $A$ on a vector space $V$ means that the direct sum of the eigenspaces $V_\lambda=\{v\in V:Av=\lambda v\}$ give the whole space $V$. (If there are non-trivial Jordan blocks, something is missing from this sum.) Then it is not hard to show that the centralizer is the direct sum/product of the full endomorphism algebras of the eigenspaces.

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