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If $A, B$ are positive $n \times n$ complex matrices, $n$ some integer, then obviously \begin{equation*} \|ABA\|_\text{tr} = tr(ABA) = tr(A^2 B). \end{equation*}

But can we say there is a constant $C_n > 0$ depending only on $n$ where $\|ABA\|_\text{tr} \geq C_n \| A^2 B\|_\text{tr}$?

Note that it's easy to get the reverse:

$\|ABA\|_\text{tr} \leq \|A^2B\| _\text{tr} $

so it's the above inequality I really need.

More generally though, I'm guessing $\|AB\|_\text{tr}$

and $\|BA\|_\text{tr}$ are not equivalent (modulo a constant depending only on $n$)

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    $\begingroup$ I don't understand what you're asking. Why do you want an inequality between two things you know are equal? $\endgroup$ Jul 26 '11 at 22:00
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    $\begingroup$ Why is $\text{tr} (A^2 B)$ necessarily $\|A^2 B\|_{\text{tr}}$ if $A^2 B$ isn't even self adjoint? $\endgroup$ Jul 26 '11 at 22:06
  • $\begingroup$ your definition of the trace norm seems to be wrong; afaik, the trace-norm is just the sum of the singular values the very first equality in your question seems to be incorrect. $\endgroup$
    – Suvrit
    Jul 26 '11 at 22:07
  • $\begingroup$ ah, ok: you are looking at just positive matrices. $\endgroup$
    – Suvrit
    Jul 26 '11 at 22:08
  • $\begingroup$ For a positive matrix $A$, the singular values ARE the eigenvalues, so isn't $\text{tr} A = \|A\|_{\text{tr}}?$ $\endgroup$ Jul 26 '11 at 22:09
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If $A$ are $B$ are projections which are not orthogonal, but are close to being orthogonal so that $ABA \not= 0$ but has only small eigenvalues then we have $\| A B A \|_\text{tr} \ll \| (A B A)^{1/2} \|_\text{tr} = \| A^2 B \|_\text{tr}$. Hence, no such constant $C_n$ exists.

For a specific example, if $A = \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right)$ and $B = \left( \begin{array}{cc} \sin^2 t & - \sin t \cos t \\ - \sin t\cos t & \cos^2 t \end{array} \right)$, then as $t \to 0$ we have $\| ABA \|_\text{tr} / \| A^2 B \|_\text{tr} = | \sin t | \to 0$.

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  • $\begingroup$ +1: very nice example. $\endgroup$
    – Suvrit
    Jul 27 '11 at 1:37
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Jesse answered your question, but let me correct your last remark. Namely if $A,B$ are positive matrices, $\|AB\|_{tr}$ and $\|BA\|_{tr}$ are more than equivalent, they are equal!

In fact more generally $AB$ and $BA$ have the same singular values. This is certainly classical, but one way to check this is to note that $(AB)^* (AB) = B A^2 B$ and $(BA)^* (BA)= A B^2 A$ have the same distribution, ie $Tr(f(B A^2 B))= Tr( f(A B^2 A))$ for any function $f:\mathbb R \to \mathbb R$. It is enough to check this equality when $f(t)=t^n$ is a monomial, in which case it is just the trace property.

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  • $\begingroup$ Thanks for answering my other question and for giving me that quick easy proof. $\endgroup$ Aug 3 '11 at 19:57

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