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Let $A, B \in M_n(\mathbb{C})$ be Hermitian. Show that $\text{tr} (AB)^2 \leq \text{tr} (A^2B^2$).

What we can say is that $AB - BA$ is skew-hermitian since $BA = (AB)^*$. Since it is Skew-hermitian, its eigenvalues are all imaginary. And its square is similar to a diagonal matrix whose diagonal entries are all negative. And so $\text{tr}(AB-BA)^2 \leq 0$.

How do you proceed from here?

ilovemeth
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  • One thing to notice, I guess, is that $\text{tr}(A^2B^2)=\text{tr}(AB\cdot BA)=\text{tr}(AB(AB)^*)=\text{tr}(|AB|^2)$. So the proof amounts to showing that this is bounded below by $\text{tr}(AB)^2=|\text{tr}(AB)|^2$. But I don't know how to proceed further. – Semiclassical Dec 07 '16 at 14:31

1 Answers1

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Just expand and use $\operatorname{tr}(XY)=\operatorname{tr}(YX)$ multiple times:

$0 \geq \operatorname{tr}(AB-BA)^2 = \operatorname{tr}(ABAB) + \operatorname{tr}(BABA) - \operatorname{tr}(ABBA) - \operatorname{tr}(BAAB) = 2\Big(\operatorname{tr}(ABAB) - \operatorname{tr}(AABB)\Big)$

This shows that $\operatorname{tr}(AB)^2 - \operatorname{tr}(A^2B^2)$ is a non-positive real number. To get the result, we still have to show that both of them are real numbers. Of course, if the difference of two numbers is real, we only have to show that one of them is real to obtain that both are real.

This is easy: $\operatorname{tr}(A^2B^2)=\operatorname{tr}(AABB)=\operatorname{tr}(ABBA)=\operatorname{tr}(AB(AB)^*) \in \mathbb R$.

MooS
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