I have been learning about matrix symmetry and came up with a question that I can't seem to prove. The idea is that the product of $ABA^T$ is a symmetric matrix.
What I mainly have to go off of is
The product of two symmetric matrices is symmetric iff the matrices commute.
I also know
$(AB)^T$ = $B^TA^T$
I know that $A$ and $B$ are symmetric matrices (not symmetric to each other) but they are not necessarily invertible matrices.
If I test an example such as
\begin{equation}
ABA^T=\begin{pmatrix}1 & 2 \\ 2 & 3\end{pmatrix}\begin{pmatrix}-4 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}1 & 2 \\ 2 & 3\end{pmatrix} = \begin{pmatrix}0 & -1 \\ -1 & -4\end{pmatrix}
\end{equation}
I get a symmetric matrix, and I can't find an example where it does not work, but I don't understand how to prove it. The reason this example is interesting to me is that $AB$ is not a symmetric matrix, but once you multiply $AB$ by $A^T$ (or just $A$ since $A$ = $A^T$), the resulting matrix is symmetrical.
But I don't know how this could be applied to make a proof that would hold for any size.
Best Answer
To show a matrix $M$ is symmetric, you just need to show that $M=M^T$.
So we want to show that $ABA^T$ is symmetric by showing that $ABA^T=(ABA^T)^T$.
Observe:
$$ (ABA^T)^T = (A^T)^T(B)^T(A)^T=AB^TA^T. $$
Since $B$ is symmetric, then $B^T=B$. Which means the equation continues as $ABA^T$. QED