# Matrix Inequality: Traces of $n$th powers

Let $$A, B$$ be matrices over $$\mathbb{C}$$ of the same dimensions (not necessarily square). With $$'$$ denoting conjugate-transpose, and tr the trace, show for $$n\in\mathbb{N}$$ that

$$2\,\mathrm{Re}\, \mathrm{tr} [(AB')^n] \le \mathrm{tr}[(AA')^n] + \mathrm{tr}[(BB')^n]$$.

As Mikael de la Salle has pointed out, this is a corollary of Hoelder's inequality and the AM-GM inequality. However, the argument below suggests that it can be viewed as coming from a combinatorial identity, as follows.

We have that RHS$$-$$LHS is a sum of squares. Write $$\|M\|^2 := \mathrm{tr}(MM')$$. As examples:

when $$n=1$$, it is $$\|A-B\|^2$$;

when $$n=2$$, it is $$\|AA'-BB'\|^2+\|AB'-BA'\|^2$$;

when $$n=3$$, it is $$\|A'BA' - B'AB'\|^2 + \frac{1}{2} \|A' A B' - B'AA'\|^2 + \frac{1}{2} \|B'B A' - A'BB'\|^2 +\\ + \frac{1}{2} \|AA'A - AB'B\|^2 + \frac{1}{2} \|AA'A - BB'A\|^2 + \frac{1}{2} \|BB'B - BA'A\|^2 + \frac{1}{2} \|BB'B - AA'B\|^2;$$

and ages ago I worked out $$n=4$$, which has coefficients 1/2, 1/3, 1/6 in it, but think I messed it up, so won't put it up here for the moment.

Anyway, by an iterative argument I can show that this is true for all $$n$$: basically I write RHS$$-$$LHS as a sum of squares plus a remainder term. I then have a procedure that makes the remainder term smaller and smaller, and I use compactness to show that in the limit the remainder goes to $$0$$.

However, that in effect uses Analysis to prove what looks like an algebraic result, and in general my expansion of RHS$$-$$LHS seems always to give rise to a sum of squares with rational coefficients, apparently of the form (integer)/$$(n-1)!$$, which my Analysis-based proof won't show. So I am now thinking that this is some sort of combinatorial identity like the Faa di Bruno formula ... What does the expansion look like for arbitrary $$n$$? Is there an explicit formula?

• Hoelder's inequality for matrices (see for example mathoverflow.net/q/78330/10265) implies that $tr((AB')^n) \leq \sqrt{$tr((AA')^n)$tr((BB')^n)}$. This is stronger than your inequality. – Mikael de la Salle Nov 8 '18 at 16:05
• To nitpick, we should better use $| \cdot |$ on the lhs at least since the matrices are allowed to be complex... – Suvrit Nov 12 '18 at 14:24
• yeah I dopily left the "Re" off - thanks. – Richard Martin Nov 12 '18 at 14:29
• @RichardMartin -- could you edit the post to highlight the precise / complete question you are asking? thanks. – Suvrit Nov 12 '18 at 20:52