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

Multiplication by ${\mathbf{v}}^{\ast }$ givesThe numerator is real because $A$ is symmetric, while the denominator is strictly positive because $B$ is positive definite. Hence $\lambda$ is real. As $A-\lambda B$ is real and singular, its kernel contains a nonzero real generalized eigenvector.