A and B commute. Delete the same row and column from each. Do they still commute?

Question

Suppose that two matrices $A$ and $B$ commute. If I delete the $i$th row and $i$th column from each, do they necessarily commute? Does the answer change if we guarantee that $A$ and $B$ are Hermitian?

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For example, consider the matrices

$A= \begin{bmatrix} \frac{4}{3} & \frac{1}{3\sqrt{2}} & -\frac{1}{\sqrt{6}} \\ \frac{1}{3\sqrt{2}} & \frac{7}{6} & -\frac{1}{2\sqrt{3}}\\ -\frac{1}{\sqrt{6}} & -\frac{1}{2\sqrt{3}} & \frac{3}{2} \end{bmatrix} $ and $B= \begin{bmatrix} \frac{5}{3} & \frac{\sqrt{2}}{3} & -\frac{\sqrt{2}}{\sqrt{3}} \\ \frac{\sqrt{2}}{3} & \frac{4}{3} & -\frac{1}{\sqrt{3}}\\ -\frac{\sqrt{2}}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & 2 \end{bmatrix} $.

A patient calculation shows that $AB - BA = 0$. Let's construct the truncated matrices $A_r$ and $B_r$ formed by deleting the third column and third row.

Then $A_r= \begin{bmatrix} \frac{4}{3} & \frac{1}{3\sqrt{2}} \\ \frac{1}{3\sqrt{2}} & \frac{7}{6} \end{bmatrix} $ and $B_r= \begin{bmatrix} \frac{5}{3} & \frac{\sqrt{2}}{3} \\ \frac{\sqrt{2}}{3} & \frac{4}{3} \end{bmatrix} $.

Another patient, but faster calculation shows that $A_r B_r - B_r A_r=0$. Then for these particular matrices, the answer to the question is yes.

A similar calculation can be done for the non-Hermitian $A= \begin{bmatrix} \frac{11}{7} & 0 & -\frac{4}{7} \\ -\frac{1}{7} & 1 & \frac{1}{7}\\ -\frac{3}{7} & 0 & \frac{10}{7} \end{bmatrix}$ and $B= \begin{bmatrix} \frac{15}{7} & 0 & -\frac{8}{7} \\ -\frac{2}{7} & 1 & \frac{2}{7}\\ -\frac{6}{7} & 0 & \frac{13}{7} \end{bmatrix}$, again affirming yes for this particular example.

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This question has a physical motivation from quantum mechanics, where whether or not two Hermitian matrices commute determines whether or not they are simultaneously observable. I'm imagining the case where we are unable to access a certain state, which would correspond to deleting the row and column corresponding to that state.

Answer 1

The answer is no in general. Consider $$ A= \begin{bmatrix} 6 & 1 & 8 \\ 7 & 5 & 3 \\ 2 & 9 & 4 \end{bmatrix} \quad\text{and}\quad B= \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} .$$ Then $AB$ computes the row sums of $A$ while $BA$ computes the column sums of $A$. Since this $A$ is a magic square (!), the answers match: $$ AB=BA= \begin{bmatrix} 15 & 15 & 15 \\ 15 & 15 & 15 \\ 15 & 15 & 15 \end{bmatrix} .$$ However, removing any row and any column breaks the magic: for example, removing the last rows and columns yields $$ A_r= \begin{bmatrix} 6 & 1 \\ 7 & 5 \end{bmatrix} \quad\text{and}\quad B_r= \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} ,$$ for which $$ A_rB_r= \begin{bmatrix} 7 & 7 \\ 12 & 12 \end{bmatrix} \quad\text{while}\quad B_rA_r= \begin{bmatrix} 13 & 6 \\ 13 & 6 \end{bmatrix} .$$

Answer 2

Without loss of generality, you may assume that the first columns and the first rows of the two matrices are deleted. Let $$ A=\pmatrix{a&u^\ast\\ v&W},\ B=\pmatrix{b&x^\ast\\ y&Z}. $$ Then $$ AB=\pmatrix{\ast&\ast\\ \ast&vx^\ast+WZ}, \ BA=\pmatrix{\ast&\ast\\ \ast&yu^\ast+ZW}. $$ The equality $AB=BA$ therefore implies that $vx^\ast+WZ=yu^\ast+ZW$. It follows that $WZ=ZW$ if and only if $vx^\ast=yu^\ast$.

In case $A$ and $B$ are Hermitian, we have $u=v$ and $x=y$. Therefore, under the assumption that $AB=BA$, the equality $WZ=ZW$ holds if and only if $ux^\ast=xu^\ast$, i.e. if and only if $x$ and $u$ are linearly dependent.