Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 5

A Hermitian matrix satisfies $H^{†}=H$, where $†$ denotes the conjugate transpose. For a nonzero eigenvector $v$ with $Hv=\lambda v$, the scalar $v^{†}Hv$ equals its own complex conjugate, andThus every Hermitian eigenvalue is real.

A unitary matrix satisfies $U^{†}U=U{U}^{†}=I$. If $Uv=\mu v$, preservation of the squared norm givesSince $v\ne 0$, $\boxed{|\mu |=1}$. These statements concern complex eigenvectors, even when the matrix entries happen to be real.

The absolutely convergent matrix exponential series givesfor Hermitian $H$. The exponents $iH$ and $-iH$ commute, so their exponentials multiply to the exponential of their sum:Hence $U$ is unitary. Alternatively, an orthonormal eigenbasis of $H$ turns the exponential eigenvalues into $e^{i{\lambda }_j}$, all on the unit circle.

Taking the adjoint of $U=M+iN$ gives $U^{†}=M-iN$. In particular, the Hermitian parts of a unitary matrix are uniquely determined:Expanding both unitary identities givesSubtracting shows $MN=NM$, and adding gives $M^2+N^2=I$. If $Mv=\mu v$, commutation implies $M(Nv)=\mu Nv$. The assumed simple spectrum makes the eigenspace one-dimensional, so $Nv=\nu v$. Thus $M,N$ have a common orthonormal eigenbasis.

On such a vector, $Uv=(\mu +i\nu )v$. If the corresponding unitary eigenvalue is $u=e^{i\theta }$, thenThe formulas for $M,N$ also show the same eigenvalue correspondence directly from $Uv=uv$ and $U^{†}v=u^{\ast }v$. Degeneracies do not prevent existence of a common basis, although the simple-spectrum assumption makes the elementary proof immediate.