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The question I would like answered is the following: Given a matrix $G$ and that $G$ commutes with another matrix $X$, that is $[G, X] = 0$, what is $X$? Or more generally, what properties of $X$ may we infer?

I understand however that this question is really too vague, so here's a more specific question: If $G$ is in Jordan canonical form, does $[G, X] = 0$ imply that $X$ has the same Jordan canonical form? Or still more specific, if $G$ is diagonal with no two diagonal entries the same, does $[G, X] = 0$ imply that $X$ is diagonal?

I have convinced myself that the answer to the latter question is ‘yes’, but a simple proof eludes me.

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  • $\begingroup$ Requiring $[G,X]=0$ will always allow $X$ to be a scalar matrix. So it is not reasonable to expect that it will ever force $X$ to have the same JNF as $G$. $\endgroup$ – Marc van Leeuwen Feb 22 '14 at 13:02
  • $\begingroup$ @MarcvanLeeuwen Yes, this is true. But can we not ask for the most general form of $X$? For example, is it not true that if $G$ is diagonal with no two entries the same then by $[G, X] = 0$ we know $X$ is diagonal also? $\endgroup$ – P. Plowman Feb 24 '14 at 9:23
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In general the question of how to describe the set of $n\times n$ matrices commuting with a given $G$ is not easy to answer. One thing one can say is that if $G$ has a minimal polynomial equal to its characteristic polynomial (a condition for which you can find a number of equivalents in the answers to this question) then any matrix commuting with $G$ is a polynomial in $G$, and this is in particular the case when $G$ has $n$ distinct eigenvalues (since the minimal polynomial must have all those eigenvalues as root). This proves a bit more than your guess that commutation of $X$ with a diagonal matrix with distinct diagonal entries forces $X$ to be diagonal.

However, when $G$ is a Jordan normal form, note that if it has multiple blocks for the same eigenvalue, then the subspaces corresponding to the blocks need not be stable under a matrix $X$ commuting with $G$, so that the latter could have a rather different Jordan normal form. Example $$ G=\begin{pmatrix} 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1\\ 0&0&0&0&0&0\\ \end{pmatrix}, \qquad X=\begin{pmatrix} 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ \end{pmatrix} $$ where $G$ has Jordan type $(3,3)$ while $X$ has Jordan type $(2,2,2)$ (and is not a polynomial in $G$).

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