MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $n$)

share|cite|improve this question
I don't understand what you're asking. Why do you want an inequality between two things you know are equal? – Qiaochu Yuan Jul 26 '11 at 22:00
Why is $\text{tr} (A^2 B)$ necessarily $\|A^2 B\|_{\text{tr}}$ if $A^2 B$ isn't even self adjoint? – Joshua Isralowitz Jul 26 '11 at 22:06
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. – Suvrit Jul 26 '11 at 22:07
ah, ok: you are looking at just positive matrices. – Suvrit Jul 26 '11 at 22:08
For a positive matrix $A$, the singular values ARE the eigenvalues, so isn't $\text{tr} A = \|A\|_{\text{tr}}?$ – Joshua Isralowitz Jul 26 '11 at 22:09
up vote 9 down vote accepted

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 $\| ABA \|_\text{tr} / \| A^2 B \|_\text{tr} = | \sin t | \to 0$.

share|cite|improve this answer
+1: very nice example. – Suvrit Jul 27 '11 at 1:37
Thanks, yea that's a nice example. – Joshua Isralowitz Aug 3 '11 at 19:57

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.

share|cite|improve this answer
Thanks for answering my other question and for giving me that quick easy proof. – Joshua Isralowitz Aug 3 '11 at 19:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.