Eigenvalues of a self-adjoint operator are real

Statement

Let $V$ be a finite-dimensional inner product space over $\R$ or $\C$, and let $T \colon V \to V$ be a self-adjoint linear map. Then each eigenvalue of $T$ is a real number.

Remarks

If $T(v) = \lambda v$ with $v \ne 0$, then $\lambda \langle v, v \rangle = \langle T(v), v \rangle = \langle v, T(v) \rangle = \bar\lambda \langle v, v \rangle$, so $\lambda = \bar\lambda$. In particular, a Hermitian matrix has only real eigenvalues. So a real symmetric matrix has only real eigenvalues, also when we view it as a complex matrix.

Explanations

L1 human

Self-adjoint maps never have complex eigenvalues that are not real. For example, the symmetric matrix $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ has the eigenvalues $1$ and $-1$. By contrast, the rotation matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ is not symmetric, and its eigenvalues are $\pm i$.