If $\langle Tv,v\rangle\in\mathbb{R}$, prove that $T$ is self-adjoint

Question

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.

If the above condition holds $\forall v\in V$, then what remains to be proved? — AlexR, Aug 05 '13 at 18:57
$T$ needs to satisfy $\left<Tv,w\right>=\left<v,Tw\right>$ for all $v,w\in V$. — John Gowers, Aug 05 '13 at 19:03
Do you know that if $\langle Tv, v \rangle = 0$ for all $v \in V$, then $T = 0$? — Ayman Hourieh, Aug 05 '13 at 19:23
Yes, There was a previous question that had us prove that. Will that be relevant here? — Makkers, Aug 05 '13 at 19:32
I added an answer. Ignore the first part if you already have a proof for it. — Ayman Hourieh, Aug 05 '13 at 19:34

score 10 · Accepted Answer · answered Aug 05 '13 at 19:34

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.

@Makkers Happy to help! – Ayman Hourieh Aug 05 '13 at 19:42 — Ayman Hourieh, Aug 05 '13 at 19:42

If $\langle Tv,v\rangle\in\mathbb{R}$, prove that $T$ is self-adjoint

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