 Solution

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.