Suppose $A\mathbf v=\lambda\mathbf v$ for a nonzero complex vector. Because a real <symmetric matrix> is <Hermitian matrix>[Hermitian], $\mathbf v^*A\mathbf v$ is real. But $\mathbf v^*A\mathbf v=\lambda\mathbf v^*\mathbf v$, 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 \b[every eigenvalue is real and has a real eigenvector].

Solution

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

 Solution (source code)