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Suppose $A,B \in {M_n}$ be Hermitian.Is this true that $tr[(AB)^2]\le tr(A^2B^2)$?

user1551
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Kavir
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1 Answers1

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Yes. Equality holds iff $AB = BA$.

Hint: Note that $AB - BA$ is skew-Hermitian, and that $$ 2\operatorname{trace}[(AB)^2] - 2\operatorname{trace}(A^2B^2) =\\ \operatorname{trace}(ABAB + BABA -ABBA - BAAB)=\\ \operatorname{trace}[(AB - BA)^2] $$

Note: The inequality assumes that both $\operatorname{trace}[(AB)^2]$ and $\operatorname{trace}(A^2B^2)$ are real. This is fine because the product of Hermitian matrices necessarily has a real trace. Note that $$ \operatorname{trace}[(A+aI)(B+bI)] =\\ \operatorname{trace}(AB) + b\operatorname{trace}(A) + a \operatorname{trace}(B) + abn $$ and that $a,b>0$ may be chosen so that $A +aI$ and $B+bI$ are positive definite. The product of (Hermitian) positive definite matrices always has positive eigenvalues.

Ben Grossmann
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    The end note is wonderful. +1 – user1551 May 30 '15 at 02:58
  • @user1551 it feels like there should be a simpler justification, but I'll settle for the quick and dirty explanation. – Ben Grossmann May 30 '15 at 03:36
  • ‘The product of positive matrices always has positive eigenvalues’ that’s interesting, can you show/hint briefly why that is true? – lcv Jul 18 '18 at 10:02
  • @icv one quick-and-dirty proof: Suppose that $A,B$ are positive. $A$ has positive definite square root $A^{1/2}$. It follows that $$ A^{-1/2}(AB)A^{1/2} = A^{1/2} B A^{1/2} $$ Since $B$ is positive definite, $MBM^*$ is positive definite for any invertible $M$, such as $M = A^{1/2}$. So, we see from the above that $AB$ is similar to a positive definite matrix. So, $AB$ has positive eigenvalues. – Ben Grossmann Jul 18 '18 at 10:37
  • [This answer](https://math.stackexchange.com/a/2884109/81360) gives a quicker proof that $AB$ has real trace. – Ben Grossmann Aug 15 '18 at 21:27