Commutant of random linear combination of matrices

Question 1

I'm not too familiar with random matrix theory so I cannot find a suitable reference for this question.

Consider a set of matrices$\{A_i\}_{i=1}^k\subseteq M_{d\times d}$ over the complex field and some random variables$\{c_i\}_{i=1}^k$, each with a (complex) Gaussian distribution.

The commutant of a set of matrices$\{X_i\}$ is defined as the set

$$ \operatorname{comm} (\{X_i\}) := \left\{ C \in M_{d \times d} \mid \left[ C,\, X_i \right] = 0, \forall i \right\} $$

Is there some reasonable definition of probability over matrices such that the statement "Consider the random matrix$A = \sum_{i=1}^k c_i A_i$, then$\operatorname{comm} (A) = \operatorname{comm} (\{A_i\}$ with high probability". Intuition and some examples tells me that this is the case, but I have not performed a large numerical study.

Is the above statement true in some reasonable setting of random matrices? Can you please provide a proof or a reference?

I would also expect an argument related to the zeroes of a polynomial to support this statement, but I have not been able to figure it out.

Question 2

Is $\operatorname{comm} (A) := \operatorname{comm} (\{A\})$?

Question 3

Yes, it is the singleton set

Question 4

This is false. A generic matrix$A$ has the property that$\text{comm}(A)$ is exactly the subalgebra$\text{span}(1, A, A^2, \dots)$ generated by$A$; a sufficient condition is that$A$ has distinct eigenvalues, in which case$\text{comm}(A)$ can also be characterized as the set of matrices with the same eigenvectors as$A$.

So you can see there will be an issue if the$A_i$ don't commute. Explicitly we can take, for example,

$$A_1 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, A_2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.$$

$\text{comm}(A_1)$ is the subalgebra of matrices of the form$\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$ while$\text{comm}(A_2)$ is the subalgebra of matrices of the form$\begin{bmatrix} a & b \\ b & a \end{bmatrix}$. Their intersection$\text{comm}(\{A_1, A_2\})$ is then the subalgebra$\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}$ generated by the identity. But$\text{comm}(c_1 A_1 + c_2 A_2)$ will always be strictly larger than this (unless$c_1 = c_2 = 0$) since it will contain the matrix$c_1 A_1 + c_2 A_2$ itself, which is never a scalar multiple of the identity (unless it's zero).

Question 5

That is why I was wondering about some sense of average/probability. The reason why I was giving that statement related more to 'sampling'. Say you first sample a certain number of matrices $A$ with different parameters $\{c_i\}$, and then compute the commutant. If I now 'take the average' of the various commutants (if this makes any sense) I would expect the 'average' commutant to be the same as that corresponding to the single $\{A_i\}$.

Question 6

@Another: I don't know what you mean by taking the "average" of the commutants.

Question 7

Yeah, me either. Intersection could work, but it is too strong of an operation. I guess the question is not well posed.

Qiaochu Yuan 477k55 gold badges1.1k silver badges1.5k bronze badges · Accepted Answer · 2025-10-25 04:30:43Z

This is false. A generic matrix$A$ has the property that$\text{comm}(A)$ is exactly the subalgebra$\text{span}(1, A, A^2, \dots)$ generated by$A$; a sufficient condition is that$A$ has distinct eigenvalues, in which case$\text{comm}(A)$ can also be characterized as the set of matrices with the same eigenvectors as$A$.

So you can see there will be an issue if the$A_i$ don't commute. Explicitly we can take, for example,

$$A_1 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, A_2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.$$

$\text{comm}(A_1)$ is the subalgebra of matrices of the form$\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$ while$\text{comm}(A_2)$ is the subalgebra of matrices of the form$\begin{bmatrix} a & b \\ b & a \end{bmatrix}$. Their intersection$\text{comm}(\{A_1, A_2\})$ is then the subalgebra$\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}$ generated by the identity. But$\text{comm}(c_1 A_1 + c_2 A_2)$ will always be strictly larger than this (unless$c_1 = c_2 = 0$) since it will contain the matrix$c_1 A_1 + c_2 A_2$ itself, which is never a scalar multiple of the identity (unless it's zero).

That is why I was wondering about some sense of average/probability. The reason why I was giving that statement related more to 'sampling'. Say you first sample a certain number of matrices $A$ with different parameters $\{c_i\}$, and then compute the commutant. If I now 'take the average' of the various commutants (if this makes any sense) I would expect the 'average' commutant to be the same as that corresponding to the single $\{A_i\}$.
@Another: I don't know what you mean by taking the "average" of the commutants.
Yeah, me either. Intersection could work, but it is too strong of an operation. I guess the question is not well posed.

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