On sum of elements in products of matrices for a simple graph

Question

Let $G$ be a simple graph with vertex set $\{v_1,v_2,\ldots,v_n\}$. The adjacency matrix of $G$ is the 0-1 matrix $A$, where $A_{i,j}=1$ when $v_i$ is adjacent with $v_j$. The degree matrix is the diagonal matrix $D$ where $D_{i,i}$ is the degree of $v_i$.

It seems that for any graph $G$, the sum of elements in any product of any $A$'s and any $D$'s (in any order) is always even. Can any one give a proof for this observation? Is it a known result? Note that the sum of elements in any power of $A$ (or $D$ ) is even, giving the first evidence to the above observation.

For example, consider $$A= \left( \begin{matrix} 0&0&1&1\\ 0&0&1&0\\ 1&1&0&1\\ 1&0&1&0\\ \end{matrix} \right), D=\left( \begin{matrix} 2&0&0&0\\ 0&1&0&0\\ 0&0&3&0\\ 0&0&0&2\\ \end{matrix} \right).$$ One can verifies that $e^TAe=8,e^TDe=8,\ldots,e^TADAAADe=422,\ldots$ are all even, where $e=(1,1,1,1)^T$.

Answer 1

I think this is false.

Here is a graph and some code in Sage to compute the sum of elements of $AADAAD$ which appears to be odd.

I tried to prove it for a while, failed, so then decided to try some randomised searching on smallish graphs.

x = Graph('G?B@dW')
A = x.adjacency_matrix()
D = matrix.diagonal([x.degree(v) for v in x.vertices()])
j = matrix([1 for i in range(x.num_verts())])
j*(A*A*D*A*A*D)*j.transpose()