Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbb K$?

share|improve this question
    
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: en.wikipedia.org/wiki/Commuting_matrices –  tomasz Sep 15 '12 at 19:55
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.