Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 5

Taking the conjugate transpose of $M{M}^{-1}=I$ and $M^{-1}M=I$ reverses the product order and givesThus $(M^{-1}{)}^{†}$ is the two-sided matrix inverse of $M^{†}$, proving

Since $A$ is Skew-Hermitian, $H=iA$ is a Hermitian matrix. The spectral theorem gives a unitary matrix $Q$ with $Q^{†}HQ=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_N)$ and real eigenvalues ${\lambda }_j$. Hence $Q^{†}AQ=\mathrm{diag}(-i{\lambda }_j)$ andIn particular the determinant is nonzero, so $I+A$ is invertible. Applying the same argument to $-A$ proves that $I-A$ is also invertible. Equality holds precisely when all ${\lambda }_j=0$, equivalently $A=0$.

Using part (i) and $A^{†}=-A$ in the Cayley transform with argument $-A$ givesThe factors commute because $I-A$ and $I+A$ are polynomials in the same matrix, and an invertible commuting matrix also commutes with the other factor's matrix inverse. Both inverses exist by part (ii). Thus $U$ is unitary.

Multiplying the defining equation on the right by $I+A$ gives $(I+U)A=I-U$. Consequently the only possible answer isThe factors commute because they are rational functions of $U$. Its conjugate transpose, using the unitarity $U^{†}=U^{-1}$, isThus $A$ is Skew-Hermitian. Moreover $I+A=2(I+U{)}^{-1}$ and $I-A=2U(I+U{)}^{-1}$, so substitution recovers $U$ and establishes existence as well as uniqueness. The unitary matrix $U=-I$ is a counterexample to invertibility of $I+U$; more generally the obstruction is an eigenvalue $-1$.