If $\langle Tv,v\rangle\in\mathbb{R}$, prove that $T$ is self-adjoint Let $V$ be a finite-dimensional vector space over $\mathbb{C}$, together with a Hermitian inner product $\langle\cdot\,,\cdot\rangle$. Let $T:V\to V$ be a linear function. Prove that $T$ is self-adjoint if $\langle Tv,v\rangle\in\mathbb{R}$.
I know that since $\langle Tv, v\rangle \in\mathbb{R}$, $\langle Tv,v\rangle=\langle v, Tv\rangle $ ... I'm not sure where to go from here.
 A: Let $S$ be a linear operator on the complex inner-product space $V$. We first show that if $\langle Sv, v \rangle = 0$ for all $v \in V$, then $S = 0$. For all $u, w \in V$, we have the identity:
\begin{align}
\langle Su, w \rangle &= \frac{\langle S(u + w), u + w \rangle - \langle S(u - w), u - w \rangle}{4} \\
&+ \frac{\langle S(u + iw), u + iw\rangle - \langle S(u - iw), u - iw\rangle}{4} i
\end{align}
This identity can be verified by direct computation. If $\langle Sv, v \rangle = 0$ for all $v \in V$, then the RHS is $0$. In the LHS, put $w = Su$ to get $\|Su\| = 0$ for all $u \in V$. Thus $S = 0$.

Now, let $T$ be a linear operator on the complex inner-product space $V$. For every $v \in V$:
\begin{align}
\langle Tv, v \rangle - \overline{\langle Tv, v \rangle} &= \langle Tv, v \rangle - \langle v, Tv \rangle \\
&= \langle Tv, v \rangle - \langle T^* v, v \rangle \\
&= \langle (T - T^*)v, v \rangle
\end{align}
Therefore, $\langle Tv, v \rangle \in \mathbb R$ if and only if $\langle (T - T^*)v, v \rangle = 0$ for all $v \in V$. But by the first part of this answer, this happens if and only if $T - T^* = 0$, if and only if $T$ is self-adjoint.
