Eigenvalues of Self-Adjoint Linear Operators

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

We will now look at a very important theorem regarding self-adjoint linear operators which says that if $T$ is self-adjoint then all eigenvalues of $T$ will be real-numbers.

Theorem 1: Let $V$ be a finite-dimensional nonzero inner product space and let $T \in \mathcal L (V)$. If $T$ is self-adjoint, then every eigenvalue of $T$ is a real number.

• Proof: Let $T$ be self-adjoint and let $\lambda \in \mathbb{F}$ be an eigenvalue of $T$. Let $v$ no a corresponding nonzero eigenvector for $\lambda$, that is, $T(v) = \lambda v$. To show that $\lambda$ is a real number, we must show that $\lambda = \bar{\lambda}$ (i.e, the imaginary part of $\lambda$ is zero).

• To show this, we note that since $T$ is self-adjoint then we have that $<T(v), v> = <v, T(v)>$ and so:

(1)

\begin{align} \quad \lambda \| v \|^2 = <\lambda v, v> = <T(v), v> = <v, T(v)> = <v, \lambda v> = \overline{\lambda} \| v \|^2 \end{align}

• Therefore $\lambda \| v \|^2 = \overline{\lambda} \| v \|^2$ which implies that $\lambda = \overline{\lambda}$ so $\lambda \in \mathbb{R}$. $\blacksquare$