Is the product of (modified) adjacency matrices of two matchings necessarily symmetric?

Question

Consider $n$ vertices, and two (not necessarily perfect) matchings $M_1$ and $M_2$. With the following definition of a (modified) adjacency matrix of a matching, can we claim that $A(M_1)\cdot A(M_2)$ is always symmetric?

Definition: A (modified) adjacency matrix $A(M)$ of a matching $M$ is an $n\times n$ matrix defined as follows: if $M$ matches a node $i$ with a node $j$, then $A(M)_{i,j} = A(M)_{j,i} = 1/2$. If $i$ is not matched with any node, then $A(M)_{i,i}=1$. All other entries are zero.

Answer 1

No, $A(M_1)A(M_2)$ need not be symmetric. Here is a counter example.
Consider the graph $K_4$ on the vertex set $[4]$.
Let $M_1=\big \{ \{1,3 \} \big \}$ and $M_2=\big \{ \{1,2 \}, \{3,4 \} \big \}$. Then $$ A(M_1)= \begin{bmatrix} 0 & 0 & \frac{1}{2} & 0 \\ 0 & 1 & 0 & 0 \\ \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 &1 \end{bmatrix} , A(M_2)= \begin{bmatrix} 0 & \frac{1}{2} & 0 & 0 \\ \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} \\ 0 & 0 & \frac{1}{2} & 0 \end{bmatrix} $$

$$ A(M_1)A(M_2)= \frac{1}{4} \begin{bmatrix} 0 & 0 & 0 & 1 \\ 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 &0 \end{bmatrix} $$ Observe that $A(M_1)A(M_2)$ is not symmetric.

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Product of two symmetric matrices $A$ and $B$ is symmetric if and only if $A$ and $B$ commute. That is, $(AB)^T=AB$ iff $AB=BA$.

Answer 2

False. Take the matching $M_1$ on four vertices that only matches $v_1$ to $v_2$, and let $M_2$ match $v_1$ to $v_3$ and $v_2$ to $v_4$. Then the matrices are $$ \begin{bmatrix} 0 & 0.5 & 0 & 0 \\ 0.5 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$

and

$$ \begin{bmatrix} 0 & 0 & 0.5 & 0 \\ 0 & 0 & 0 & 0.5 \\ 0.5 & 0 & 0 & 0 \\ 0 & 0.5 & 0 & 0 \\ \end{bmatrix} $$

whose product is $$ \begin{bmatrix} 0 & 0 & 0 & 0.25 \\ 0 & 0 & 0.25 & 0 \\ 0.5 & 0 & 0 & 0 \\ 0 & 0.5 & 0 & 0 \\ \end{bmatrix} $$

I thought about these matrices as $\frac{1}{2}$ the adjacency matrix of your matching with two loops at isolated vertices. Then the product of matrices is $\frac{1}{4}$ times the matrix whose entries are paths from $i$ to $j$ of length $2$ going first along an edge in the first graph then along an edge in the second graph. After first thinking that this was obviously symmetric, drawing out an example ( this one ) I saw it was not.