Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 6

Let $A=A^{†}$ be a Hermitian matrix and $Av=\lambda v$ with $v\ne 0$. The scalar $v^{†}Av$ equals its own complex conjugate, becauseSince $v^{†}Av=\lambda v^{†}v$ and the squared norm $v^{†}v$ is positive, $\boxed{\lambda \in \mathbb{R}}$.

Now take $Au=\lambda u$ and $Av=\mu v$. The inner product satisfiesSubtracting gives $(\mu -\lambda )u^{†}v=0$. Thus distinct eigenvalues imply $\boxed{u^{†}v=0}$, proving orthogonality of the corresponding eigenvectors. Within a degenerate eigenspace, an orthonormal basis may be chosen, but arbitrary distinct vectors in that same eigenspace need not be orthogonal.

A complex square matrix $U$ is a unitary matrix if $\boxed{U^{†}U=U{U}^{†}=I}$. Equivalently, it preserves the complex inner product and hence the norm. If $Uv=\mu v$ with $v\ne 0$, thenDivision by the positive squared norm gives $\boxed{|\mu |=1}$. Thus every eigenvalue lies on the unit circle.

Part (a) makes each ${\lambda }_j$ real. In the eigenvector basis,and $|{\lambda }_j+ia{|}^2={\lambda }_j^2+a^2>0$. Hence $\boxed{A+iaI\text{ is invertible}}$.

All factors that are rational functions of $A$ commute. Taking the Hermitian conjugate reverses product order and givesThereforeso $\boxed{V\text{ is unitary}}$. This is a Cayley transform of a Hermitian matrix, with scale $a$ and the convention that an infinite real eigenvalue maps to one.

Applying it to each eigenvector yieldsThe eigenvectors are unchanged, and $|{\nu }_j|=1$ because numerator and denominator are complex conjugates. No ${\nu }_j$ equals one when $a>0$. The scalar map is injective on real $\lambda$, so the eigenspaces, including any degeneracies, are preserved exactly.

Use $U^{†}=U^{-1}$ and take the Hermitian conjugate in the reverse product order:The factorization $I-U^{-1}=-U^{-1}(I-U)$ gives $(I-U^{-1}{)}^{-1}=-(I-U{)}^{-1}U$, while $I+U^{-1}=U^{-1}(I+U)$. ConsequentlyThe last equality uses the fact that $I+U$ commutes with $I-U$ and its inverse. Thus $B$ is a Hermitian matrix. The exclusion of the eigenvalue one is essential to define the inverse. Spectrally, an eigenvalue $e^{i\vartheta }\ne 1$ is sent to $-\cot (\vartheta /2)$, a real number. This is the inverse Cayley transform of a unitary matrix without eigenvalue one.