Help With the Proof of the Spectral Decomposition (Quantum Mechanics) $\newcommand\dag\dagger$
I need to present a proof of the spectral decomposition and I need help in some parts. I will state the theorem and the proof indicating where help is needed. I know this is more mathematical than physics but I encountered it doing Quantum Mechanics. Any help would be much appreciated.
Theorem: Any normal operator $M$ on a vector space $V$ is diagonal with respect to some orthonormal basis for $V$ 
Proof: Let dimension of vector space $V$ be $d$. We will use the second principle of Mathematical Induction:
1. For $d=1$: A one dimensional matrix over the complex field is just a complex number. The adjoint of a complex number is also a complex number and since $2$ complex numbers commute, the matrix is normal. But it is also clear that it is diagonal for any orthonormal basis in V. Hence true for $d=1$.
2. Assume $d<n$: Assume that a normal operator on a vector space $V$ of dimension $d$ which is less than $n$ is diagonal with respect to some orthonormal basis for $V$.      
3. For $d=n$: By the fundamental Theorem of Algebra, the characteristic equation for $M$ has at least one solution and hence $M$ has at least one eigenvalue. Let one of the eigenvalues of $M$ be $\lambda$. 
Let $P$ be the projector onto the $\lambda$ eigenspace. This means that for any arbitrary vector $|\psi\rangle$,
$$MP |\psi\rangle = \lambda P |\psi\rangle\qquad (*)$$ 
Let $Q$ be the projector onto the orthogonal complement satisfying $PQ=0$. Since $P$ and $Q$ are projectors the satisfy: $$P^\dag=P, Q^\dag=Q, P^2=P, Q^2=Q $$
Now, here is my first question: The proof says $P+Q=I$. Is this how we define an orthogonal compliment or can it be derived from some other equation? 
Using $P+Q=I$,  $M$ can be written as:
\begin{aligned}
M &= (P+Q)M(P+Q) \\
& = PMP+QMQ+QMP+PMQ \\
\end{aligned}
Using $(*)$ and $QP=O$ we have that $QMP=0$
Now, to show $PMQ=0$:
Let $|v\rangle$ be an arbitrary eigenvector belonging to the subspace $P$. Then: $$MM^\dag|v\rangle=M^\dag M|v\rangle=M^\dag (\lambda |v\rangle)=\lambda M^\dag |v\rangle$$
This means that $M^\dag |v\rangle$ also belongs in $P$ subspace. Hence $QM^\dag P=0$. My second question is: Is this the case since for any arbitrary vector $|\psi\rangle$, $P|\psi\rangle$ 'lands' in the $P$ subspace, say to a vector $|v\rangle$. Then $M^\dag |v\rangle$ remains in the subspace and it follows $QM^\dag P=0$ since $Q$ is orthogonal to any element of $P$ subspace by definition? 
From $QM^\dag P=0$, $(QM^\dag P)^\dag =0$ which implies $PMQ=0$. Thus we have reduced $M$ to $$M=PMP+QMQ$$
Now we show $QMQ$ is normal. To do this we first derive two results: $$QM=QM(P+Q)=QMQ$$ and $$QM^\dag =QM^\dag (P+Q)= QM^\dag Q$$
Using these two results we have: 
\begin{aligned}
QMQ(QMQ)^\dag &= QMQQM^\dag Q\\
& = QMQM^\dag Q \\
& = QMM^\dag Q \\
& = QM^\dag MQ \\
& = QM^\dag QMQ \\
& = (QM^\dag Q)(QMQ)
\end{aligned}
Hence $QMQ$ is normal. 
This is the main part I don't understand. The proof continues as and I quote: 
"By induction $QMQ$ is diagonal with respect to some orthonormal basis for the subspace of Q because it's dimension is definetely less than k as the rest of dimensions is occupied by P subspace (whose dimension is at least 1).
 Using $(*)$, $PMP$ is diagonal in its basis. Thus using Mathematical Induction, we conclude that $M=PMP+QMQ$ is diagonal with respect to some orthonormal basis for the total vector space." 
I don't understand how we use a basis of $Q$ to diagonalize $QMQ$ and similarly how we use a basis of $P$ to diagonalize $PMP$. Also I don't understand how $M=PMP+QMQ$, doesn't this imply that $dim(PMP)=dim(QMQ)$? Otherwise how could we add the components of each matrix. I guess these questions arise because I don't understand what $PMP$ and $QMQ$ represent. 
 A: First question: Basically  yes. The orthogonal complement is the kernel of $P$, the set of vectors $|w\rangle$ such that $P|w\rangle = 0$. Then for any such $|w\rangle$ it holds that $(I - P)|w\rangle \equiv Q |w\rangle = |w\rangle$, while for any $|v\rangle$ such that $P|v\rangle = |v\rangle$, it follows that $(I - P)|v\rangle \equiv Q |v\rangle = 0$.
Second question: Yes again, correct.
Main question: Note that the subspace projected by $Q$ necessarily is of dimension $k<dim(V)=n$. But according to assumption 2 any normal operator on a (sub)space of dimension $d<n$ is known to be diagonal with respect to some basis in that (sub)space. Apply this to $QMQ$ wrt the subspace of $Q$ on which it is defined and the statement that "QMQ is diagonal with respect to some orthonormal basis for the subspace of $Q$" follows necessarily. A similar statement holds for $PMP$, also a normal operator, wrt the subspace of P, because $dim(P) < n $ as well. So $PMP$ must have a diagonal representation on the eigenspace projected by $P$. Since both $QMQ$ and $PMP$ have diagonal representations, so does $M = PMP + QMQ$.
