Diagonalizable Matrix Intuition: $P^2=P$. $\textbf{Definition:}$ Let $P$ be a matrix. We say is diagonalizable matrix iff 

$\exists$ an invertible matrix $A$: $\exists$ a diagonal matrix $D$:  $A^{-1}PA=D$.

I recently asked a question as to why the following proposition holds true:
$\textbf{Proposition:}$ Let $P$ be a $n\times n$ matrix.

If $P^2=P$, then $P$ is diagonalizable.

$\textbf{Question:}$ Let $v\in \mathbb{R}^n$. I am trying to better understand this proposition (see below):

$\textbf{Why}$ do we rewrite $v=v-Pv+Pv$? It strikes me as an obvious statement which holds true, but it is always rewritten for a particular reason. And $\textbf{what}$ is the general idea as to why eigenvalues are roots of the following expression, $Q(P):=P^2-P$? I thought the eigenvalues might have been associated with the Cayley-Hamilton theorem the way things are set up, but I could be wrong.

Roots for $Q(P)$,
$P^2-P=0\implies P(P-1)=0\implies 1 \text{ and } 0 \text{ are Eigenvalues}$.


 A: I'll try to give some more details as an answer.
From the relation $\,P^2=P$, you obtain that $\;P(I-P)=0$, so $\operatorname{Im}(I-P)\subset \ker P$, and conversely, you can check $\ker P\subset \operatorname{Im}(I-P)$. Thus   $\ker P=\operatorname{Im}(I-P)$, and if $P\ne I$, this image is non-zero so you have the eigenspace for the eigenvalue $0$.
Now $P^2-P$ corresponds to the polynomial $X^2-X$, which is the minimal polynomial of $P$ if $P\ne 0,I$. As the eigenvalues are the roots of the minimal polynomial, you know there are no other eigenvalue than $0$ and $1$.
Further, $\operatorname{Im}(P)=\ker(I-P)$ is clearly the eigenspace for the eigenvalue $1$.
A: Let $P$ be so that $P^2=P$. Let $z \in \mathbb R^n$. Define
$$z_1=P(z) \\
z_2=z-z_1$$
Then, 
$$P(z_1)=z_1=1 \cdot z_1 \\
P(z_2)=0=0\cdot z_1\\
z=z_1+z_2$$
This shows that each vector can be written as linear combination of eigenvectors for $P$. Deduce from here that, if $E_0,E_1$ are the eigenspaces, then 
$$\dim(E_1)+\dim(E_0)=n$$
and hence $P$ is diagonalizable.
