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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.

Tobias
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Charles Chou
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  • What is the matrix $A'$ in relation to $A$? – wgrenard Nov 16 '16 at 01:11
  • Suppose we have the special case: $B + B' = C^{2}$ for some self-adjoint operator $C$. Then $$ \mbox{tr }(A C^2 A') = \mbox{tr }(A C C' A') = \mbox{the Frobenius norm of $AC$}, $$ and the latter norm overestimates $\mbox{tr }(A B A')$ approximately (and sometimes exactly) by a factor of $2$. – avs Nov 16 '16 at 01:15
  • Thanks, I am looking for lower and upper bound for the trace. – Charles Chou Nov 16 '16 at 18:59

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One bound you might like (using the Cauchy Schwartz inequality on the Hilbert Schmidt inner product) is $$ |tr(ABA')| = |tr(A'AB)| \leq tr(A'A)\sqrt{tr(B'B)} $$

Ben Grossmann
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