An exercise I have come upon while studying Hilbert Spaces:
Let $A$ be a compact operator, and $P \in L(H)$ be an orthoprojection. Prove that $$\lambda_n (PA^*AP) \leq \lambda_n (A^*A)$$ (Where $\lambda$'s are the eigenvalues, and the indexing is from the minimax principle, i.e. $\lambda_1$ is the biggest eigenvalue, and so on.)
There's also a hint:
Define $B = A^*A$, then $B$ is compact self-adjoint. Then $$Bx = \sum_n \lambda_n \langle x,\phi_n \rangle \phi_n$$ $$PBPx = \sum_n \lambda_n \langle x,P\phi_n \rangle P\phi_n$$ "...and from there the proof is trivial."
But I do not see the triviality of how to continue. The biggest problem for me, conceptually, is that I can not manage to separate the corresponding eigenvalues to compare between the two, they always come all together in a sum.
I do not ask for the complete proof with all the details, but perhaps someone could outline the idea of the proof in general? Thanks.
Recall the min-max theorem for compact operators, $$\lambda_n(B)=\mbox{max }_{\mbox{dim }V=n}\mbox{min }_{x\in V,\|x\|=1}\langle Bx,x\rangle.$$ If $\lambda_n(PBP)=0$, there is nothing to prove, so assume $\lambda_n(PBP)>0$. Pick an $n$-dimensional subspace $V$ such that $$\lambda_n(PBP)=\mbox{min }_{x\in V,\|x\|=1}\langle BPx,Px\rangle.$$ Now, since $\lambda_n(PBP)>0$, we have $\|Px\|>0$ for all $x\in V$, and thus \begin{align*} \mbox{min }_{x\in V,\|x\|=1}\langle BPx,Px\rangle&= \mbox{min }_{x\in V,\|x\|=1}\|Px\|^2\langle BPx/ \|Px\|,Px/\|Px\|\rangle\\ & \leq \mbox{min }_{x\in PV,\|x\|=1}\langle Bx,x\rangle. \end{align*} Again, since $\|Px\|>0$, it follows that $\dim PV=\dim V$, and therefore $$\lambda_n(PBP)=\mbox{min }_{x\in V,\|x\|=1}\langle Bx,x\rangle\leq \mbox{max }_{\mbox{dim }V=n}\mbox{min }_{x\in V,\|x\|=1}\langle Bx,x\rangle=\lambda_n(B).$$