How does this proof for spectral decomposition work?

Question

My book has a proof for spectral decomposition that proceeds as follows: Let M be a normal operator, let λ be an eigenvalue of M, let P be the projector onto the λ eigenspace and let Q be be the projector onto the orthogonal complement. Then M = (P + Q)M(P + Q) = PMP + QMP + PMQ + QMQ, the proof continues by proving that QMP = 0 and PMQ = 0. I can prove that QMP = 0 easily, but proving that PMQ = 0 has proven difficult.

The proof in the book is: let v be a member of the λ eigenspace, then $M^†Mv = MM^† v = \lambda M^† v$, therefore $QM^†P=0$ and from that PMQ=0, but this doesn't make sense to me. That only applies when v is a member of the λ eigenspace and not when it's an arbitrary vector.

I've taken a picture of the proof in the book to make sure that I'm not misreading it.

Answer 1

In $T$ is a normal operator, then $\|T^*x\|=\|Tx\|$. Therefore, $\mathcal{N}(T^*)=\mathcal{N}(T)$.

Your $M$ is normal, as is $M-\lambda I$. Therefore \begin{align} (M-\lambda I)P=0 &\implies (M^*-\overline{\lambda}I)P=0 \\ &\implies P(M-\lambda I)=0 \\ &\implies P(M-\lambda I)Q=0 \\ &\implies PMQ=0. \end{align}

Answer 2

As you say: $$M^†Mv = MM^† v = \lambda M^† v$$ which means $$MM^† v = M(M^†v)=\lambda(M^†v)$$

Therefore $M^†v$ is in the $\lambda$ eigenspace or here the $P$ subspace. Hence $Q(M^†P) = 0$ and $PMQ = 0$ follows from that by taking the conjugate transpose.