Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 1 / 8C / b / Solution

A complex square matrix $U$ is a unitary matrix when $U^{†}U=I$, equivalently $U^{-1}=U^{†}$. Here $†$ denotes the adjoint matrix, the conjugate transpose. Such a matrix preserves the standard complex inner product and its Euclidean norm.

A real symmetric matrix $A$ is a Hermitian matrix, so $A^{†}=A$. For arbitrary complex $\mathbf{x}$,The imaginary cross terms cancel because $A$ is Hermitian. Reversing the signs gives the same identity:In particular, $(A+iI)\mathbf{x}=0$ implies $|\mathbf{x}{|}^2=0$, so $A+iI$ has zero kernel of a linear map. It is a square matrix, and the rank-nullity theorem therefore makes it invertible. The same argument proves invertibility of $A-iI$.

The proposed matrixis consequently well-defined. For any $\mathbf{y}$, put $\mathbf{x}=(A+iI{)}^{-1}\mathbf{y}$. The two norm identities show directly thatTo verify the unitary matrix identity explicitly, take adjoints:The two shifts commute, because both are polynomials in $A$. HenceThusThe imaginary shifts are invertible, and their equal norm changes cancel in the quotient. This is the Cayley transform of a Hermitian matrix in the $(A-iI)(A+iI{)}^{-1}$ convention.