Past exam of the mathematics course of the University of Cambridge / 2005 / ib / Paper 4 / 1B (Linear Algebra)

Use the standard complex inner product $⟨u,v⟩=u^{\ast }v$, conjugate-linear in its first argument. A unitary matrix satisfies $U^{\ast }U=I$, and a Hermitian matrix satisfies $H^{\ast }=H$, where $\ast$ is conjugate transpose.

If $Hv=\lambda v$ with $v\ne 0$, thenPositivity of $⟨v,v⟩$ proves $\boxed{\lambda \in \mathbb{R}}$. If $Uv=\lambda v$, preservation of the norm gives $\Vert v{\Vert }^2=\Vert Uv{\Vert }^2=|\lambda {|}^2\Vert v{\Vert }^2$, so $\boxed{|\lambda |=1}$.

For eigenvectors $Hv=\lambda v$ and $Hw=\mu w$, the Hermitian matrix identity gives $⟨v,Hw⟩=⟨Hv,w⟩$. The already established reality of the eigenvalues therefore implies $(\mu -\lambda )⟨v,w⟩=0$. Thus distinct eigenvalues have orthogonal eigenvectors.