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What is the dimension of the subspace of $M_n(\mathbb C)$ consisting of matrices $B$ such that $AB = BA$, for some fixed matrix $A$ in $M_n(\mathbb C)$?

user192906
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    Do you know what $A$ is? The answer will change depending on whether $A = I_n$ or $A$ is a random matrix. – bzc Nov 16 '14 at 06:37

1 Answers1

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It depends on the Jordan canonical form of $A$. For example, suppose

$$ A = \pmatrix{\lambda & 1 & 0 & 0 & 0\cr 0 & \lambda & 1 & 0 & 0\cr 0 & 0 & \lambda & 0 & 0\cr 0 & 0 & 0 & \lambda & 1\cr 0 & 0 & 0 & 0 & \lambda\cr}$$

Then in order to commute with $A$, $B$ must be of the form

$$ \left( \begin {array}{ccccc} a&b&c&d&e\\ 0&a&b&0&d \\ 0&0&a&0&0\\ 0&f&g&h&i \\ 0&0&f&0&h\end {array} \right) $$ In this case there are $9$ free parameters, so the space is $9$-dimensional.

In general, suppose $A$ has $m$ Jordan blocks of sizes $s_i$, $i= 1.. m$, and let $e(i,j) = 1$ if blocks $i$ and $j$ have the same eigenvalue, $0$ otherwise. Then the dimension is $$\sum_{i=1}^m \sum_{j=1}^m e(i,j) \min(s_i,s_j)$$

Robert Israel
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