Let $P\in L(X)$ be a projection operator, where $X$ is a complex non-trivial banach space, that is $P^2=P$ and $Q\in L(X)$ such that $Q^2=0$, then in my Functional analysis exam it was asked to see that we will have $(P+Q)^2=P+Q$, however I cannot seem to prove this , we wil have that $(P+Q)(P+Q)=P^2+PQ+QP+Q^2=P+PQ+QP$ but how can I relate $PQ+QP$ with $Q$?
I also tried coming up with a counterexample if we consider $X=l^2(\mathbb{N})$ and the operator $P(x)=\langle x,e_1\rangle e_1$ and the operator $Q(x)=Q((x_1,x_2,x_3,...)=(0,x_1,0,x_3,0,x_5,...)$ isn't this going to be a counterexample? Because we will have that $(P+Q)(x)=(x_1,x_1,0,x_3,0,...)$ and that $(P+Q)^2(x)=(x_1,x_1,0,0,...)$.
Thanks in advance.
Your counterexample works !
My idea:
For $P=I$ we do not have $(P+Q)^2=P+Q$ for each $Q$ with $Q^2=0:$
$(I+Q)^2=I+2Q$, thus
$$(I+Q)^2=I+Q \iff I+2Q=I+Q \iff Q=0.$$
Someone told you nonsense !