Linear Algebra - Proving a projection given a linear transformation

Question

Suppose that $V$ is a vector space, and $M$ is a subspace of $V$.

A transformation $P:V \rightarrow V$ is called the projection of $V$ onto $M$ if

(i) there exists a subspace $N$ such that every vector $v \in V$ can be written uniquely as $v = x + y$ for some $x \in M$ and $y \in N$; and

(ii) $P$ is given by $P(x + y) = x$, for all $x \in M$ and $y \in N$.

Suppose that $P:V \rightarrow V$ is a linear transformation. Prove that $P$ is a projection onto some subspace of $V$ if and only if $P^2 = P$.

How do I prove that the linear transformation of $P$ projects onto a subspace of $V$ iff $P^2 = P$

Answer 1

Take $M=P(V)$ and $N=\ker P$. Then, for each $v\in V$,$$v=\overbrace{v-P(v)}^{\in N}+\overbrace{P(v)}^{\in M}.$$Note that $v-P(v)\in\ker P$ because$$P\bigl(v-P(v)\bigr)=P(v)-P^2(v)=P(v)-P(v)=0.$$On the other hance, if $v=x+y$, with $x\in M$ and $y\in N$, then $x=P(x')$ for same vector $x'$So,$$P(v)=P(x)+P(y)=P^2(x')=P(x')=x$$and$$y=v-x=v-P(v).$$

Answer 2

Hint:

$\Rightarrow $

$P^2(v)=P^2(x+y)=P(x)=P(v)\,,\forall v$

$\Leftarrow $

Let $M=\operatorname{im}P$ and $N=\operatorname{ker}P$.