Let $X$ be a Banach space and let $\Lambda\in\mathfrak L(X,X)$ be compact s. t. $\Lambda=\Lambda ^2$. Namely, $$\Lambda(x)=\Lambda\left(\Lambda(x)\right),\;\forall x\in X.$$ Prove that $\mathcal Im(\Lambda)$ is finite-dimensional.
Let's take a bounded set $A\subset X$ s.t. $\Lambda(A) = B_1$, so $\Lambda (A)$ is a unit ball in the $\mathcal Im(\Lambda)$ and by the compactness of the operator it is pre-compact, hence closed and bounded.
Applying the operator again,$\Lambda(\Lambda(A))$ is also pre-compact (closed and bounded).
Since $\Lambda\left(\Lambda(A)\right)=\Lambda(A)$ , then $\Lambda(\Lambda(A)) = B_1$ is also a unit ball, so $B_1$ is a pre-compact unit ball. So $\mathcal Im(\Lambda)=\mathcal Im(\Lambda^2)$ is finite dimensional.
May I ask for verification and advice on improving my proof if necessary?