What commutes with a matrix in Jordan canonical form? 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.
 A: 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 $G$).
