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Let $A$ be an $n\times n$ complex matrix, and $C(A)$ be the vector space of all matrices that commute with $A$. I have to determinate if the dimension of $C(A)$ is greater or equal than $n$, or not.

If anyone can give me a hint, i think the answer is yes, but i am not sure what i have to use to prove it.

Dimitri
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    For starters, what must be true of the size of matrices in $C(A)$? If $B\in C(a)$ then $BA=AB$, so we know that we can multiply $B$ on both the right and left of $A$. – Eoin Aug 14 '14 at 05:59
  • If $A$ has (pairwise) distinct eigenvalues then $C(A)$ consists of polynomials evaluated at $A$, and consequently $\dim C(A)=n$. If $A$ has repeated eigenvalues, the dimension is higher if $A$ is also diagonalizable. I think that double centralizer theorems have a lot to say about this. – Jyrki Lahtonen Aug 14 '14 at 06:18
  • I was thinking something related with minimal polynomial of $A$, because i noticed that the dimension of $C(A)$ seems to be lower when the degree of the minimal is greater, and in the case that the degree is n, the dimension is greater or equal that n as you said, but i am not sure if i am on the right way – Dimitri Aug 14 '14 at 06:42

2 Answers2

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For any matrix $A$ , $\dim \mathcal{C}(A)\geq n$, moreover if $A$ is diagonalizable with $\lambda_1,\ldots,\lambda_r$ its eigenvalues and $n_i=\text{multi}(\lambda_i)$, then $\dim\mathcal{C}(A)=\displaystyle\sum_{i=1}^rn_i^2$.

Hamou
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Hint: consider the Jordan canonical form of $A$. What can you say about matrices that commute with a Jordan block?

Robert Israel
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