Past exam of the mathematical methods I course of the University of Oxford / 2022 / 10

For unitary matrices, $U^{†}U=I$ and $V^{†}V=I$. The adjoint matrix reverses the order of multiplication, soThus the product is unitary. Likewise $(UV)(UV{)}^{†}=I$, or this follows from invertibility in finite dimension.

Take the adjoint matrix of $M^{†}{\mathbf{v}}_i={\lambda }_i{\mathbf{v}}_i$ to obtain ${\mathbf{v}}_i^{†}M={\lambda }_i^{\ast }{\mathbf{v}}_i^{†}$. Evaluating ${\mathbf{v}}_i^{†}M{\mathbf{u}}_j$ using this equation and $M{\mathbf{u}}_j={\lambda }_j{\mathbf{u}}_j$ givesHence $\boxed{{\mathbf{v}}_i^{†}{\mathbf{u}}_j=0\text{ if }{\lambda }_i^{\ast }\ne {\lambda }_j}$. The proof does not assume that $M$ is a normal matrix or diagonalizable.

In general the spectrum of an adjoint matrix is the complex conjugate of the original spectrum, not necessarily the same set. The equality of the two sets in the question is an additional assumption, possible for a conjugation-invariant spectrum; the displayed orthogonality proof only uses the two stated eigenvalue equations.

A column of coordinates transforms as ${\mathbf{x}}^{\prime }=U^{†}\mathbf{x}$, because the columns of $U$ are the new orthonormal basis vectors expressed in the old coordinates. A symmetric unitary choice isIts first and third columns are orthogonal eigenvectors of the $xz$ block, with eigenvalues $-1$ and $+1$ respectively; its second column is the $y$ axis with eigenvalue $3$. Direct multiplication checks $U^{†}U=I$, while the displayed entries give $U^T=U$. The symmetry here means transpose symmetry, not Hermitian symmetry.