Conjecture that $A^{T}BA = ABA^{T}$ for any symmetric matrix $B$ in $\mathbb{R}^n$

Question

While trying to understand the Kalman filter, and by experimentation with Python I came up with the conjecture in the title.

First of all is it true? Second, if it is, how can I prove this? I would prefer a more elegant, short proof than an elementary, one, so don't avoid e.g. bilinear forms or the diagonalisability of symmetric matrices, if these things elucidate why it is true, and make the proof shorter.

Answer 1

Not it's not. For example, take $B=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $A=\begin{pmatrix}1&0\\1&1\end{pmatrix}$. Then $$ A^TBA = \begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}1&0\\1&1\end{pmatrix} = \begin{pmatrix}2&1\\1&1\end{pmatrix} $$ but $$ ABA^T = \begin{pmatrix}1&0\\1&1\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix} = \begin{pmatrix}1&1\\1&2\end{pmatrix} $$