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Suppose I have a matrix $A $ in the space $ V $ of $n $ by $n $ matrices. Then it is quite clear that $S=\{B : AB=BA\} $ form a subspace. I want to find out its dimension. I think it depends on the rank of $A $.

I'm trying the simplest case : $A $ has full rank. If $X$ is commutative with any other elements of $V $, then $X $ belongs to $S$. It is only the multyple of the identity matrix. So $1\leq \text {dim} S $.

But It's hard for me to make any other proper upper bound.

MrTanorus
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  • The simplest case is when $A=0$. I don't know the answer to this, but I suspect it's related to the pattern of zeros in $A$, though this is clearly insufficient by itself. Look at https://yutsumura.com/basis-for-subspace-consisting-of-matrices-commute-with-a-given-diagonal-matrix/ for a start. – saulspatz Aug 17 '19 at 15:02
  • @saulspatz That problem suggest that If $A=\text{daig}(a_1,...,a_1, a_2, ....,a_2, ..., a_n, ... ,a_n)$ while $a_i$ appears $d_i$ times, then $\text{dim} S= \sum {a_i ^2}$. So it gives full answer for diagonalizable $A$! – MrTanorus Aug 17 '19 at 16:05

1 Answers1

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The rank of $A$ has nothing to do with it. For a cheap counter-example, consider that the zero matrix (rank 0) and the identity matrix (full rank) both commute with every matrix and hence $S$ would have dimension $n^2$ for both.

You may find it useful to know that two matrices commute if and only if they preserve each other's eigenspaces. If you think about this a bit harder (or a lot harder and know some Lie algebra theory), you might come to some of the following....

Commuting Matrices Burnside's Theorem Lie-Kolchin Theorem

erfink
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