Suppose $0\leq P\leq I$ is some positive semidefinite matrix. Here $A\leq B$ implies that $B-A$ is positive semidefinite.
For a positive semidefinite $X$, can the largest eigenvalue of $PXP$ be larger than the largest eigenvalue of $X$? If not, how can one show this?
My proof idea:
Let $x$ be the normalized eigenvector of $X$ corresponding to the largest eigenvalue. Then $x^*Xx \geq y^*Xy$ for any normalized $y$. In particular, $y^*PXPy = zXz$, where $z = Py$. Since $y^*y = 1$, we have $z^*z = yP^2y \leq y^* I y = 1$. So $\|z\|_2 \leq 1$. From this, we conclude that $x^*Xx \geq y^*PXPy$ for all normalized $y$ and hence the maximum eigenvalue cannot increase.
I did not use the positive semidefiniteness of $X$ in the proof and this is concerning because a projection can increase the maximum eigenvalue of a negative definite matrix. So I think something is wrong with the proof which leaves the original question still open.
Best Answer
You have $x^*Xx \geq y^*Xy$ for all $x$ and $y$ normalized vector. But on your proof $z = Py$ can be $0$ which is non-normalizable. A concrete example take $X = \begin{pmatrix} -1 & 0 \\ 0 & -2 \end{pmatrix}$. $e_1$ is the eigenvector corresponding to $-1$ the largest eigenvalue. Then taking $P=0$ we have $PXP=0$ with largest eigen value $0$ and $z=P e_1 =0$ is not "normalizable".