if $T$ is normal and a projection in a finite dimensional vector space $V$, then $T$ is an orthogonal projection Let $T$ be a normal operator ($TT^* = T^*T$) defined in a finite dimensional vector space $V$ with inner product ($<v, w>$). Then, if $T$ is a projection ($T^2 = T$) then $T$ is an orthogonal projection ($N(T)^\bot = R(T)$ and $R(T)^\bot = N(T)$).
where $N(T)$ is the Kernel of $T$ and $R(T)$ is the Range or Image of $T$.
Attempt:
by taking $y \in R(T)$ and $w\in N(T)$ I'm trying to prove that $N(T)^\bot = R(T)$ . I must prove $\langle  y,w\rangle  = 0$, which means that $N(T) = R(T)^\bot$
$\langle y,w\rangle = \langle T(x),w\rangle   =  \langle x, T^*(w)\rangle $ Then I dont know what to do. Also
$\langle y,w\rangle  = \langle  T(x),w\rangle  = \langle T^2(x), w\rangle $ Then I don't know what to do.
Also, I know that $T$ is an orthogonal projection if and only if $T$ is autoadjunct ($T^*=T$)
Therefore, I could try to prove that $T$ is autoadjunct. But I don't know if this is even true. Any hint?

As V is finitely dimensional then is sufficient to prove $\mathcal N(T) = \mathcal R(T)^\bot  $
let be $x \in \mathcal R(T)^\bot$ then
$0 = <x, \ T(T^*(x))>\ =\ <x,\ T^*(T(x))>\ =\ <T(x),\ T(x)>$
i.e  $\ <T(x),\ T(x)> = 0 \iff T(x) = 0 \iff x \in \mathcal N(T)$
Therefore $R(T)^\bot \subseteq N(T)$
To check $N(T) \subseteq R(T)^\bot $ is the same backwards
let be $x \in \mathcal N(T)$ then
$0= <T(x),\ T(x)> = <x,\ T^*(T(x))> = <x,\ T(T^*(x))>$
Therefore $<x,\ T(T^*(x))> = 0$ with  $T(T^*(x)) \in R(T) \implies x \in \mathcal R(T)^\bot$
Therefore $N(T) \subseteq R(T)^\bot $
Then $N(T) = R(T)^\bot $
Thus $T$ is an orthogonal projection
