# 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 $
= and so:$

(1)

\begin{align} \quad \lambda \| v \|^2 = = = = = \overline{\lambda} \| v \|^2 \end{align}

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