Let $T\in \mathcal{L}(V)$ be an operator such that $T^{2}=T$. Prove the following are equivalent:
a) $T$ is self-adjoint
b) $T$ is normal
c) $v-Tv$ is orthogonal to $Tv$ for all $v \in V$
Attempt:
$\bullet a) \Rightarrow b)$ $T^{2}=T$ is the definition of idempotent operator. Since $T$ is idempotent and self-adjoint, $TT^{*}=T^{2}=T^{*}T$.
$\bullet b) \Rightarrow a)$ A normal, idempotent linear operator must be self-adjoint
So I have $a) \Leftrightarrow b)$ but I don't know how to get c) in there.
$b) \implies c)$
Assume $T$ is normal. This implies
$$\text{Im}(T)^\perp=\ker(T^*)=\ker(T)=\text{Im}(I-T).$$
Therefore, $\left \langle (I-T)v|Tw \right \rangle=0,\forall v,w \in V$. If $v=w$, $v-Tv$ is orthogonal to $Tv$.
$c) \implies a)$
Assume $v-Tv$ is orthogonal to $Tv$ for all $v\in V$. This means
$$0=\left \langle Tv|v-Tv \right \rangle=\left \langle Tv|(I-T)v \right \rangle=\left \langle ((I-T)^* T)v|v \right \rangle.$$
If $V$ is a finite $\mathbb{C}$-space and $P\in \mathcal{L}(V)$ is such that $\left \langle Pv|v \right \rangle=0,\forall v\in V$, then $P=0$. This implies
$$0=(I-T)^* T=T-T^*T.$$
That is, $T=T^*T$.
$$T^*=(T^*T)^*=T^*(T^*)^*=T^*T=T,$$
so $T$ is self-adjoint.