# Is a projection operator diagonal-decreasing on a positive operator?

I just faced a obvious-looking inequality, but I didn't manage to prove it. Let $H$ be a finite-dimensional Hilbert space, $M, \rho$ positive operators on $H$, $P$ an orthogonal projector on $H$. Is it true, that $\text{tr}(M \rho) \geq \text{tr}(M P \rho P)$ ? If so, how can one prove that? If not, what would be a counterexample?

A somewhat weaker, but sufficient condition would be that $\langle M v, v \rangle \geq \langle PMPv, v \rangle \quad \forall v \in H$, but I still can't prove it.

Thanks for any help.

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The weaker condition is true. Indeed, call $S$ the image of $P$. Then every vector $v \in H$ can be decomposed uniquely as $v=Pv+v_{\bot}$ with $v_{\bot}\in S^\bot$ and $(Mv, v)=(PMPv, v)+(Mv_\bot, v_\bot)$. Since $M$ is positive we have $(Mv_\bot, v_\bot) \ge 0$. – Giuseppe Negro Sep 27 '11 at 8:20
What about the cross-terms? $(Mv, v) = (M(Pv + v_\bot), Pv + v_\bot) = (MPv,Pv) + (Mv_\bot, v_\bot) + (Mv_\bot, Pv) + (MPv, v_\bot)$. I don't see why the cross-terms should be non-negative. – Tom Jonathan Sep 27 '11 at 8:46
DANG! You're right, sorry about that. What I said is true only if $S$ reduces $M$. – Giuseppe Negro Sep 27 '11 at 13:29

M = (9 -2; -2 1) P = (1 0; 0 0 ) rho = (9 2; 2 1)

Yes, it is indeed. It is also a counterexample to the weaker statement, taking $v = (1;1)$. – Nate Eldredge Sep 27 '11 at 14:45