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Martin Sleziak
Martin Sleziak
7:47 PM
For instructions how to render MathJax(TeX) in chat see this post on meta or go directly to robjohn's website.
1
A: For non-diagonalizable matrices, the dimension of centralizer can be different from $\sum\limits_{j=1}^k d_j^2$

Martin SleziakIf $A$ consists of a single Jordan block of size $n\times n$, i.e., $$A=J= \begin{pmatrix} \lambda & 1 & \ldots & \ldots & 0 \\ 0 & \lambda & 1 & & \vdots \\ \vdots & & \ddots & \ddots & 0 \\ \vdots & & & \lambda & 1 \\ 0 & \ldots & \ldots & 0 & \lambda \\ \end{pmatrix}, $$ then we...

This all checks out. More generally, we will have $\dim C(A) \geq n$ if and only if the matrix $A$ is non-derogatory, i.e. if and only if the Jordan form of $A$ has one Jordan block for each eigenvalue. — Omnomnomnom 6 hours ago
More generally, I believe that the result for a matrix with one eigenvalue $\lambda$ will be as follows: if $d_k$ denotes the number of Jordan blocks with size $k$, then $\dim C(A) = \sum_k k\cdot d_k^2$ — Omnomnomnom 6 hours ago
That first comment should say "we will have $\dim C(A) = n$". We will always $\dim C(A) \geq n$. — Omnomnomnom 6 hours ago
Actually for $d_1=d_2=2$ I get $\dim C(A)=5$.
For $J=
\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
$ and $X=
\begin{pmatrix}
a & b & c \\
0 & a & 0 \\
0 & d & e \\
\end{pmatrix}
$ I get
$$\begin{align*}
JX&=
\begin{pmatrix}
0 & a & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}\\
XJ&=
\begin{pmatrix}
0 & a & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
\end{align*}$$
If I had to make a guess, if $n_1,\ldots,n_s$ are dimensions of the blocks, I would guess that $$\dim C(A)=\sum_{i=1}^s \sum_{j=1}^s \min{n_i,n_j}.$$
@Omnomnomnom Actually, I think that for $d_1=d_2=1$ I get $\dim C(A)=5$. I left a few comments in my chatroom. — Martin Sleziak 15 secs ago
 
Martin Sleziak
Martin Sleziak
8:06 PM
I suppose that if the case for single eigenvalue could be solve somehow, this could help to get to the general case: Commuting block diagonal matrices.
Here is the guess once again, without the missing brackets: $$\dim C(A)=\sum_{i=1}^s \sum_{j=1}^s \min\{n_i,n_j\}.$$
 
Martin Sleziak
Martin Sleziak
8:19 PM
This could be probably the same as $\sum_k c_k^2$ if $c_k$ denotes the number of blocks with the size at least $k$.
 

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