Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 1 / 7B (Vectors and Matrices)

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.

From $A\mathbf{v}=\lambda B\mathbf{v}$ we have $(A-\lambda B)\mathbf{v}=0$ with $\mathbf{v}\ne 0$. Thus $A-\lambda B$ is singular and $\det (A-\lambda B)=0$.

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.

For $A\mathbf{u}=\lambda B\mathbf{u}$ and $A\mathbf{v}=\mu B\mathbf{v}$, symmetry givesIf $\lambda \ne \mu$, ${\mathbf{u}}^{T}B\mathbf{v}=0$: the eigenvectors are orthogonal in the inner product induced by $B$.

The generalized characteristic polynomial isThe generalized Rayleigh quotient is maximized by the largest generalized eigenvalue, $4$. Solving $(A-4B)\mathbf{v}=0$ gives