Idempotent Operators If $P$ is an idempotent operator, $P^2 = P$ and we have a vector $|\psi\rangle$, $P\neq1$, and the relation
$$PL |\psi\rangle = L|\psi\rangle,$$ what conclusions can we draw about $L$, which is a Linear Operator?


*

*$L = P$.


Are there anything else? If not how does one prove this?
 A: Without further constraints you cannot say anything about $f$. Observe that the identity is idempotent and $id \circ f = f$ holds for any function $f$ on any set.
Alright. If $P$ is not the identity, it cannot be invertible, thus $\ker(P) \neq 0$. Idempotent operators are basically projection operators, they are diagonalizable and only have eigenvalues $0$ or $1$ (this is not hard to prove, but tell me if you wish me to). Now, if $P L |\psi\rangle = L |\psi\rangle$, this simply means that $\mathrm{im}(L) \subset \mathrm{im}(P)$, i.e. the operator $L$ takes only values in the subspace $P$ projects onto. I do not see that anything more can be said.
Note that the above assumes that your equation holds for all $|\psi\rangle$. You say "we have a vector", but from one vector alone one could not deduce anything (except that $\mathrm{im}(L) \cap \mathrm{im}(P) \neq 0$).
EDIT
Proof of diagonalizability:
If $P^2 = P$, then $\mathrm{im}(P)$ is an eigenspace of weight $1$ of $P$, since $P|\psi\rangle = |\psi\rangle \forall |\psi\rangle \in \mathrm{im}(P)$, so $P$ is the identity on its image. Now, by basic linear algebra, $V = \mathrm{im}(P) \oplus \mathrm{ker}(P)$ (this is the first isomorphism theorem or splitting lemma). Since $P$ is the identity on its image and the zero map on its kernel (which are both obviously diagonal), it is diagonal on their direct sum.
The statement "If $L$ is idempotent, then $P = L$" follows now from doing the same argument for $L$ and observing that they are both the identity on $\mathrm{im}(P)$ and the zero map on the kernel, thus they are the same operators.
EDIT2:
And this is what happens when you don't think about what you write: We cannot conclude that $P = L$ even if $L$ is idempotent, since it could be the projection on a subspace of $\mathrm{im}(P)$ and not on $\mathrm{im}(P)$ itself. We would need $LP|\psi\rangle = P|\psi\rangle$ for that, too.
A: Sometimes it's helpful to consider finite dimensional special case to get insights on a problems like these.  Let $P$ and $L$ be in $\mathbb{C}^{n \times n}$. If $P^2 = P$, then $Pv = \lambda v$ leads to $\lambda^2 = \lambda$, i.e., the eigenvalues are in $\{0, 1\}$.
In the context of quantum, we consider Hermitian operators. In this case, we can find some basis so that $$P = \begin{bmatrix}I & 0 \\ 0 & 0\end{bmatrix}, \quad L = \begin{bmatrix}A & B \\ B^\dagger & C\end{bmatrix}, \quad
\therefore PL = \begin{bmatrix}A &B \\ 0 & 0\end{bmatrix}.$$
where $A$ and $C$ are Hermitian.  Now, given $PL = L$, we get $B = 0$ and $C = 0$, thus arriving at
$$L = \begin{bmatrix}A & 0 \\ 0 & 0\end{bmatrix}.$$
In particular, this means $P$ and $L$ commute, and $Ker(P) \subseteq Ker(L)$.
For general linear operators, we start with $P^\dagger = P$, $L^\dagger = L$, $P^2 = P$ and $PL = L$. $Ker(P) \subseteq Ker(L)$ is trivial.  Meanwhile,
$$LP = (P^\dagger L^\dagger)^\dagger = (PL)^\dagger = L^\dagger = L = PL,$$
i.e., commutativity holds in general.
