Orthogonal Projection; Range and Kernel I am trying to prove the following

Let $M$ be a subspace of $V$, prove that
(1) $R(P_M) = M$ and (2) $Ker(P_M)=M^\perp$ where $P_M$ is the projection onto $M$

For (1) I have
Proof
$R(P_M) = \{ m \in V: m = P_M(m) \}$ and $M = \{v \in V : v \in M \}$
Let $x \in R(P_M),$ then $x  = P_M(x) \implies x \in M$ and $x \in M \implies x = P_M(x)$
For (2) I have
Proof
$Ker (P_M) = \{ m \in M: P_M(m) = 0 \}$ and $M^\perp = \{ v \in V: \langle v, m\rangle = 0, \forall m \in M \}$
Let $x \in Ker(P_M) \implies P_M(x) = 0 $. I am not sure what to do now.

Given vectors $u,v$ we may write $u=u_1+u_2$, such that $u_1$ is a multiple of $v$, and $\langle u_2,v\rangle=0$, i.e. $u$ is normal to $v$.  $u_1$ is called the projection of $u$ onto $v$, and is given by $\frac{\langle u,v\rangle}{\langle v,v\rangle}v$.  $u_2$ is found via $u_2=u-u_1$.  Projection onto $M$ is defined in a different way, but yields the same decomposition.
Now, if $x\in M$, then when we write $x=x_1+x_2$, $x_2=0$.  If $x\in M^\perp$, then when we write $x=x_1+x_2$, $x_1=0$.  Proving these two facts (using the definition of projection onto $M$ you were given) will solve this problem.
