Self-Adjoint Linear Operators over Complex Vector Spaces

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 $<T(v), v> = 0$ for all $v \in V$ then $T$ is identically equal to the zero operator.

 Proposition 1: If $V$ is a complex inner product space and $T \in \mathcal L (V)$ is such that $= 0$ for all vectors $v \in V$, then $T = 0$.
• Proof: Let $V$ be a complex inner product space. Then for all $u, w \in V$ we can write $<T(u), w>$ as:
(1)
\begin{align} \quad <T(u), w> = \frac{<T(u+w), u + w> - <T(u - w), u - w>}{4} + \frac{<T(u + iw), u + iw> - <T(u - iw), u - iw>}{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 <T(u), w> = \frac{<T(v_1), v_1> - <T(v_2), v_2>}{4} + \frac{<T(v_3), v_3> - <T(v_4), v_4>}{4} \end{align}
• Now suppose that $<T(v), v> = 0$ for all vectors $v \in V$. Then the righthand side of the equation above reduces to zero and $<T(v), w> = 0$ 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 $\in \mathbb{R}$ for all vectors $v \in V$.
• Proof: $\Rightarrow$ Let $V$ be a complex inner product space and let $v \in V$. Suppose that $<T(v), v> \in \mathbb{R}$ for all vectors $v \in V$. Then $<T(v), v> = \overline{<T(v), v>}$ and so:
(3)
\begin{align} \quad 0 = <T(v), v> - \overline{<T(v), v>} \\ \quad 0 = <T(v), v> - <v, T(v)> \\ \quad 0 = <T(v), v> - <T^*(v), v> \\ \quad 0 = <(T - T^*)(v), v> \end{align}
• Therefore $<(T - T^*)(v), v> = 0$ 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 <T(v) - v> - \overline{<T(v), v>} = <(T - T^*)(v), v> \\ \quad <T(v) - v> - \overline{<T(v), v>} = <0(v), v> \\ \quad <T(v) - v> - \overline{<T(v), v>} = 0 \\ \end{align}
• Therefore $<T(v), v> = \overline{<T(v), v>}$ which implies that $<T(v), v> \in \mathbb{R}$ for all $v \in V$. $\blacksquare$