Suppose $A$ and $B$ are square matrices, is there any simple way to find the
trace or the bound of the trace of $ABA^T$?
Thanks.
$B$ has the form of \begin{align*} \begin{pmatrix} 0 & a_{1} & a_{2} & a_{3} \\ 0 & 0 & a_{1} & a_{2} \\ 0 & 0 & 0 & a_{1} \\ 0 & 0 & 0 & 0 \end{pmatrix}. \end{align*} $A$ is unrestricted.
One bound you might like (using the Cauchy Schwartz inequality on the Hilbert Schmidt inner product) is $$ |tr(ABA')| = |tr(A'AB)| \leq tr(A'A)\sqrt{tr(B'B)} $$