Properties of symmetric matrix with row/column sums equal to 0.

Question

If $A$ is $p\times n$ matrix and $B$ is $p\times p$ symmetric matrix with row/column sums equal to $0$. Is it true that $A^TBA$ will have row/column sums equal to $0$ for any $A$? Also, is $A^TBA$ symmetric for any $A$?

Answer 1

Let A be $\begin{bmatrix} 1&2 \\ 3&4\end{bmatrix}$, B be $\begin{bmatrix} 1&-1 \\ -1&1\end{bmatrix}$. $A^TBA = \begin{bmatrix} 4&4 \\ 4&4\end{bmatrix}$ So it is definitely not the case that the row/column sums to 0 for any A.

Now, consider the product $C=A^TBA$ , $C_{ij} = i_{th}$ row of $A^T$ multiplied by $j_{th}$ column of $BA$. Now since $A(BC)$ = $(AB)C$, $C_{ji} = j_{th}$ row of $A^TB$ multiplied by $i_{th}$ column of $A$.

Hence, I will show that $C_{ij}=C_{ji}$, since $C_{ij}$ is an entry, taking the transpose of it would yield the same result(the value itself). Note, $Row_i(A^T)*Col_j(BA)$ = $Col_i(A)*Col_j(BA)$

Thus, taking the transpose of $Col_i(A)*Col_j(BA)$ we have $(Col_j(BA))^T$$Col_i(A)^T$, which is equal to $(Col_j(BA))$$Col_i(A)$.
Note previously we had $C_{ji}$ = $Row_j(A^TB)Col_i(A) = Col_j(A^TB)^TCol_i(A)=Col_j(B^TA)Col_i(A)$. Since B is a symmetric matrix, this is equal to $(Col_j(BA))$*$Col_i(A)$. The indices $i,j$ were chosen to be arbitrary, so the product is then symmetric.