Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 1 / 7B / a

Suppose $A\mathbf{v}=\lambda \mathbf{v}$ for a nonzero complex vector. Because a real symmetric matrix is Hermitian, ${\mathbf{v}}^{\ast }A\mathbf{v}$ is real. But ${\mathbf{v}}^{\ast }A\mathbf{v}=\lambda {\mathbf{v}}^{\ast }\mathbf{v}$, so $\lambda$ is real. The real matrix $A-\lambda I$ is singular; the real or imaginary part of a nonzero complex null vector supplies a nonzero real eigenvector. Thus every eigenvalue is real and has a real eigenvector.

If $A\mathbf{u}=\lambda \mathbf{u}$ and $A\mathbf{v}=\mu \mathbf{v}$, symmetry givesFor $\lambda \ne \mu$, ${\mathbf{u}}^T\mathbf{v}=0$, so the eigenvectors are orthogonal.

Eigenvectors belonging to distinct eigenvalues are linearly independent. With $n$ distinct eigenvalues there are $n$ eigenvectors forming a basis, and the matrix is diagonalizable. In this eigenbasis, the Rayleigh quotient is a weighted average of the eigenvalues, so it is maximized by choosing $\mathbf{v}$ in the eigenspace of the largest eigenvalue.