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


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