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?

Question

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!

Answer 1

$$\langle Tv,v\rangle=\overline{\langle Tv, v\rangle}=\langle v, Tv\rangle$$

The first equality follows because $\langle Tv,v\rangle\in\mathbb{R}$ and the second by antisymmetry of the scalar product.

Answer 2

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!!!