PREMISE: Suppose you are given a matrix $S$ and you have to find a projector $P$ (which is a hermitian matrix that satisfies $P^2=P$) that is a solution of $$PSP=\lambda(S)P,$$ for $\lambda(S)$ some complex number.
QUESTION: Can I manipulate this expression and make it look like a conventional eigenvalue problem? By this I mean something in the form:$$M(S)Q(P)=\lambda(S)Q(P),$$ where $Q(P)$ is a column vector and $M(S)$ is a matrix. Then I would simply have to get the matrix $P$ back from the solution $Q$.
WHAT I THOUGHT OF SO FAR: If I think of $P$ as a `vector of rows' and write $$(PS)P_i=\lambda(S)P_i\quad\forall i,$$ where $P_i$ is the $i$-th row of $P$, it kind of looks like I'm searching for the eigenvectors of $PS$, but this is almost no progress, because $P$ is still in $PS$... Any idea would be greatly appreciated!
If you find an eigenvector $v$ for $S$ with eigenvalue $\lambda$, you can then take $P$ to be the projection onto the span of $v$: concretely, $$ P=\frac1{(v^*v)^{1/2}}\,vv^*. $$ This works, because if $Sv=\lambda v$, then for any $x$ you have $SPx=\lambda Px$, and so $PSPx=Px$. As this holds for any $x$, you have $PSP=\lambda P$.