Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2003 / Paper 1 / 5

If $Hv=\lambda v$ for a Hermitian matrix and a nonzero eigenvector, then $v^{†}Hv=\lambda v^{†}v$. The scalar on the left equals its complex conjugate because $H^{†}=H$, while $v^{†}v>0$. Thus $\lambda$ is real. If also $Hw=\mu w$, thenConsequently distinct eigenvalues force $v^{†}w=0$, proving orthogonality of their eigenvectors.

For a unitary matrix, $U^{†}U=I$ preserves the norm induced by the standard Hermitian inner product. If $Uv=\zeta v$ with $v\ne 0$, then $\Vert v{\Vert }^2=\Vert Uv{\Vert }^2=|\zeta {|}^2\Vert v{\Vert }^2$, so $\boxed{|\zeta |=1}$.

Since $A$ and $B$ are real symmetric matrices, $U^{†}=A-iB$. Expanding the unitarity condition givesEquality of the entrywise real and imaginary parts yieldsThese are the identities for the real and imaginary parts of a symmetric unitary matrix.

Let $Av=\alpha v$. The commutation relation gives $A(Bv)=B(Av)=\alpha Bv$, so $B$ preserves the eigenspace of $A$ with eigenvalue $\alpha$. Because it is a nondegenerate eigenvalue, the eigenspace is one-dimensional, and $Bv=\beta v$ for some real $\beta$. This includes $Bv=0$. Thus the eigenvectors of $A$ form a common eigenbasis, andWriting the corresponding unit-modulus eigenvalue of $U$ as $\zeta =e^{i\theta }$ givesThe pairing refers to the same common eigenvector; independent reorderings of the two real spectra would lose this information. The simultaneous diagonalization can also be constructed without requiring simple eigenvalues by diagonalizing $B$ within each eigenspace of $A$.