Let V be a complex vector space. If $\langle T(v),v\rangle\in \mathbb R$ for every $v \in V$, then is T self adjoint? Let V be a complex vector space. If $\langle T(v),v\rangle\in \mathbb R$ for every $v \in V$, then is T self adjoint? 
I know how to prove reverse claim. Not sure about this one. 
Thanks! 
 A: $$\langle Tv,v\rangle=\overline{\langle Tv, v\rangle}=\langle v, Tv\rangle$$
The first equality follows because $\langle Tv,v\rangle\in\mathbb{R}$ and the second by antisymmetry of the scalar product.
A: It is true, at least if $V$ is of finite dimension, as the following argument demonstrates:
Since $\langle Tv, v \rangle \in \Bbb R$ for all $v \in V$, we have
$\langle v, Tv \rangle = \langle Tv, v \rangle^* = \langle Tv, v \rangle, \tag{1}$
where ${}^*$ denotes complex conjugation, and I have adopted the simplified notation $Tv$ for $T(v)$; thus $Tv = T(v)$ for all $v \in V$.  From (1), 
$\langle T^\dagger v, v \rangle = \langle Tv, v \rangle, \tag{2}$
whence
$\langle (T^\dagger - T)v, v \rangle = 0. \tag{3}$
Observe that $T^\dagger - T$ is skew-adjoint, viz.
$(T^\dagger -T)^\dagger = T - T^\dagger, \tag{4}$
from which it follows that $i(T^\dagger - T)$ is self-adjoint, since
$(i(T^\dagger - T))^\dagger = -i(T -T^\dagger) = i(T^\dagger - T). \tag{5}$
Since $i(T^\dagger - T)$ is self-adjoint, it may be diagonalized over $\Bbb C$, and its eigenvalues are real.  But if 
$i(T^\dagger - T)e = \mu e \tag{6}$
then (3) becomes
$0 =  \langle i(T^\dagger -T)e, e \rangle = \mu \langle e, e \rangle, \tag{7}$
implying $\mu = 0$ for all eigenvalues $\mu$; this in turn forces 
$T^\dagger - T = 0, \tag{8}$
that is, $T$ is self-adjoint. QED.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
