Let $T$ be linear operator on a finite dimensional inner product space $V$ such that $T^2=T$. Determine whether $\ker T=\operatorname{Range}\,(T)^\perp$.
I have proved that $\ker T=\operatorname{Range}\,(T)^\perp$ under the assumption that $T$ is Hermitian. I guessed that the answer is yes but still in trouble to make it. I also want to know whether $T^2=T$ on $V$ will imply that $T$ is Hermitian.
No. Consider the example on $\mathbb C^2$ of
$$T = \left( \begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix} \right)$$
Then $T^2 = T$, $T$ is not Hermitian and $\ker T = \text{span} \left\{ \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) \right\}$ which is not orthogonal to $\text{range}(T) = \text{span} \left\{ \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) \right\} $
In fact, $\ker T = ({\cal R} T)^\bot$ iff $T$ is Hermitian.
Note that $\ker T = {\cal R} (I-T)$.
Suppose $\ker T = ({\cal R} T)^\bot$. Then $\langle (I-T)x, Ty \rangle = 0$ for all $x,y$, hence $(I-T^*)T = 0$, or $T= T^* T$, and so $T$ is Hermitian.
Now suppose $T$ is Hermitian.
If $x \in \ker T$, then $\langle x, Tz \rangle = \langle Tx,z \rangle = 0$ for all $z$ and so $x \in ({\cal R} T)^\bot$.
If $x \in ({\cal R} T)^\bot$, then $\langle x, Tz \rangle =\langle Tx,z \rangle = 0$ for all $z$ and so $Tx = 0$, hence $x \in \ker T$.
