Eigenspaces of a self-adjoint operator are orthogonal

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. Let $v$ and $w$ be eigenvectors of $T$ for distinct eigenvalues $\lambda \ne \mu$. Then $v$ and $w$ are orthogonal. So the eigenspaces of $T$ for distinct eigenvalues are orthogonal to each other.

Remarks

The proof: $\lambda \langle v, w \rangle = \langle T(v), w \rangle = \langle v, T(w) \rangle = \mu \langle v, w \rangle$, since $\mu$ is real. So $(\lambda - \mu) \langle v, w \rangle = 0$, and $\langle v, w \rangle = 0$.

Explanations

L1 human

For a symmetric matrix, eigenvectors for different eigenvalues are perpendicular. So the eigenspaces meet at right angles. This is the first step towards an orthonormal basis of eigenvectors.