# Self-Adjoint Linear Operators over Complex Vector Spaces

Recall from the Self-Adjoint Linear Operators page that if $V$ is a finite-dimensional nonzero inner product space and if $T \in \mathcal L (V)$ then $T$ is said to be self-adjoint if $T = T^*$.

In the following proposition we will see that if $V$ is a *complex* inner product space, that if $

Proposition 1: If $V$ is a complex inner product space and $T \in \mathcal L (V)$ is such that $ |

**Proof:**Let $V$ be a complex inner product space. Then for all $u, w \in V$ we can write $$ as:

(1)

\begin{align} \quad = \frac{ – }{4} + \frac{ – }{4} \end{align}

- Let $v_1 = u + w$, $v_2 = u – w$, $v_3 = u + iw$ and $v_4 = u – iw$. Then the equation above can be rewritten as:

(2)

\begin{align} \quad = \frac{ – }{4} + \frac{ – }{4} \end{align}

- Now suppose that $
= 0$ for all vectors $v \in V$. Then the righthand side of the equation above reduces to zero and $= 0$ for all $u, w \in V$ which implies that $T = 0$. $\blacksquare$

With this proposition, we will see in the next corollary that if $V$ is complex inner product space then $T$ will be self-adjoint if and only if the inner product between $v$ and its image $T(v)$ is zero for all vectors $v \in V$.

Corollary 1: If $V$ is a complex inner product space and $T \in \mathcal L (V)$ then $T$ is self-adjoint if and only if $ |

**Proof:**$\Rightarrow$ Let $V$ be a complex inner product space and let $v \in V$. Suppose that $\in \mathbb{R}$ for all vectors $v \in V$. Then $= \overline{ and so:}$

(3)

\begin{align} \quad 0 = – \overline{} \\ \quad 0 = – \\ \quad 0 = – \\ \quad 0 = \end{align}

- Therefore $ = 0$ for all $v \in V$. By Proposition 1, this implies that $T – T^* = 0$ and so $T = T^*$, that is, $T$ is self-adjoint.

- $\Leftarrow$ Suppose that $T$ is self-adjoint. Then $T = T^*$. From above we still have that:

(4)

\begin{align} \quad – \overline{} = \\ \quad – \overline{} = \\ \quad – \overline{} = 0 \\ \end{align}

- Therefore $
= \overline{ which implies that $}$ \in \mathbb{R}$ for all $v \in V$. $\blacksquare$