Proof of this trace inequality

Question

In an article I was reading, it is written: It is well-know that for any two matrices X and Y of appropriate sizes we have $$|tr\, X^*Y| \leq \frac{tr \, X^*X + tr \, Y^*Y}{2}$$

I didn’t find it obvious and I couldn’t find a proof. I’ve tried to it it terms of summation but did work. How can I prove it? Or where can I find a proof?

Answer 1

The map $(X,Y)\mapsto tr(X^*Y)$ is a positive definite inner product on the set of matrices. Hence, by Cauchy-Schwarz, $$ |\operatorname{tr}(X^*Y)|\le \sqrt{\operatorname{tr}(X^*X)\operatorname{tr}(Y^*Y)}. $$ Now use the inequality $2\sqrt a\sqrt b\le a+b$, which holds for all $a,b\ge 0$ by the binomial formula.

Answer 2

Hint: Use $\operatorname{tr}(A+B) = \operatorname{tr}A+\operatorname{tr}B$ and \begin{align} \operatorname{tr}((X-Y)^\ast(X-Y))\geq 0. \end{align}

Additional Hint: Also consider \begin{align} \operatorname{tr}((X+iY)^\ast(X+iY))\geq 0. \end{align}