The context of the problem I have is that $T$ is supposed to be an idempotent/projection operator. (Actually I'm trying to re-understand an answer[at 1. fifth line "If..."] I got about a month before)
About my current understanding: I'm really confused about that given an idempotent operator $T^2=T$, and combine with $T=T^*$ then it becomes an orthogonal projection operator.
I know a little about four fundamental subspaces and some properties about adjoint operator. But I didn't connect the dots to understand what happen when $T=T^*$ will make the $N(T)$ and $R(T)$ "orthogonal".
Since I don't fully understand the relation so I didn't assume $T$ to be idempotent in the following original proof and it's incorrect...
(For $N(T)$ means the null space of $T$, while $R(T)$ means the range of $T$.)
Let $T$ be a linear operator defined on an finite inner product space $V$. What I want to prove is that
\begin{align*} R(T)=R(T^*)\implies T=T^* \end{align*}
so $T$ is self-adjoint.
Assume $R(T)=R(T^*)$. Since
$$N(T)^\perp=R(T^*)=R(T)\\ R(T)^\perp=N(T^*)=R(T^*)$$
and by the fact that $V=W\oplus W^\perp$ where $W$ is any subspace of $V$
$$V=N(T)\oplus R(T)\\ V=N(T^*)\oplus R(T^*)$$
and $N(T^*)=R(T)^\perp=R(T^*)^\perp=N(T)$, so
$$N(T)=N(T^*)$$
but does this mean $T=T^*$? Please help...
From the decomposition $V=N(T)\oplus R(T)$, and $N(T)=N(T^*)$ and $R(T)=R(T^*)$, you have enough to finish.
Given $x\in V$, write $x=v+w$, where $v\in N(T)$ and $w\in R(T)$. There exist $u,u'\in V$ such that $w=Tu=T^*u'$. We have $$Tx=T(v+w)=Tu=T^*u'=T^*(v+T^*u')=T^*(v+w)=T^*x.$$ Since $x\in V$ was arbitrary, $T=T^*$.