Dimension of the space of matrices which commutes to an gaussian random matrix

Question

Let $$ be an $×$ matrix with elements randomly sampled from standard normal distribution $N(0,1)$. Let $()$ denote the set of all matrices which commute with $$.

What is the dimension of $()$?

Answer 1

As I note in my comment, a matrix $A$ with elements that are sampled from a normal distribution has distinct (complex) eigenvalues with probability $1$.

First of all, note that when I mean that $A$ has distinct eigenvalues, I mean there are no repeated complex eigenvalues (that is, it is not sufficient to restrict our focus to real eigenvalues). For one argument that most matrices have distinct eigenvalues, see this post. As that post demonstrates, the set of matrices having a repeated eigenvalue has Lebesgue measure $0$. Because of the absolute continuity of the normal distribution (as a probability measure), we can conclude that the probability of $A$ having a repeated eigenvalue is $0$, which is to say that $A$ has distinct eigenvalues with probability $1$.

As for the dimension of $C(A)$ assuming that $A$ has distinct eigenvalues, we can proceed as follows. First, we consider the simpler case of matrices with complex entries. That is, we consider $$ C_{\mathbb C}(A) = \{X \in \Bbb C^{n \times n} : AX = XA\} $$ We will consider the dimension of this subspace of $\Bbb C^{n \times n}$ as a vector space over $\Bbb C$.

1. Note that if $S^{-1}AS = D$, then $X \mapsto S^{-1}XS$ is a bijective linear map that takes $C_{\mathbb C}(A)$ to $C_{\mathbb C}(D)$. It follows that $\dim C_{\mathbb C}(A) = \dim C_{\mathbb C}(D)$.

2. Show that if $D$ is diagonal with distinct diagonal entries, then $DX = XD$ if and only if $X$ is also diagonal. Conclude that $\dim C(D) = n$.

3. Conclude that if $A = S^{-1}DS$ and $D$ has distinct diagonal entries, then $\dim C_{\mathbb C}(A) = \dim C_{\mathbb C}(D) = n$.

4. Reframe the result as follows: the linear map $\Phi: \Bbb C^{n \times n} \to \Bbb C^{n \times n}$ given by $\Phi(X) = AX - XA$ has a kernel $C_{\mathbb C}(A)$, which has dimension $n$. Notably, $A$ is a matrix with real entries.

5. Use this to conclude (perhaps with the help of this post) that the linear map $\Phi:\Bbb R^{n \times n} \to \Bbb R^{n \times n}$ given by $\Phi(X) = AX - XA$ also has a kernel of dimension $n$. That is, $\dim C(A) = n$.