# If A and B are orthogonal projection matrices, how can I show that trace(AB) <= rank(AB)?

If A and B are orthogonal projection matrices, how can I show that trace(AB)≤ rank(AB)?
I was using C-S inequality to get tr$\left(AB\right)\le \sqrt{tr\left({A}^{2}\right)tr\left({B}^{2}\right)}$ and I know that $tr\left({A}^{2}\right)=$rank(A). But I can't get the rank of AB.
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Jaelyn Levine
If either A or B is zero, this holds trivially.
Suppose that both A and B are non-zero. It suffices to show that all eigenvalues of AB have magnitude at most equal to 1. To that end, note that if $‖\cdot ‖$ denotes the spectral norm, then we have $‖A‖=‖B‖=1,$ so that $‖AB‖\le ‖A‖\cdot ‖B‖=1$. It follows that all eigenvalues $\lambda$ of AB satisfy $|\lambda |\le ‖AB‖\le 1$. Thus, if AB has rank r and ${\lambda }_{1},\dots ,{\lambda }_{k}$ (with $k\le r$) are the non-zero eigenvalues of AB, then we have
$\mathrm{tr}\left(AB\right)\le |\mathrm{tr}\left(AB\right)|=|\sum _{i=1}^{k}{\lambda }_{i}|\le \sum _{i=1}^{k}|{\lambda }_{i}|\le \sum _{i=1}^{k}1=k\le r,$
which is what we wanted.