 Solution

Use the complex inner product $⟨a,b⟩={\sum }_j\overline{a_j}{b}_j=a^{†}b$, conjugate-linear in its first argument. For $a\ne 0$, put $c=⟨a,b⟩/⟨a,a⟩$. Positivity givesHence the Cauchy-Schwarz inequality is $|⟨a,b⟩|\le \Vert a\Vert \Vert b\Vert$. If $a=0$ it is immediate; equality holds precisely when the two vectors are linearly dependent.

For a unitary matrix, $U^{†}U=I$, so $\Vert Ua{\Vert }^2=a^{†}{U}^{†}Ua=\Vert a{\Vert }^2$. If $Uv=\lambda v$ with a nonzero eigenvector, this implies $|\lambda |\Vert v\Vert =\Vert v\Vert$. Thus every eigenvalue satisfiesFor the one-parameter family, expand unitarity to first order: $(I+t{A}^{†}+O(t^2))(I+tA+O(t^2))=I$, so $A^{†}+A=0$. Thus $A$ is a skew-Hermitian matrix, an element of the unitary Lie algebra. If $Av=\mu v$, then $v^{†}Av=\mu \Vert v{\Vert }^2$, whereas complex conjugation gives $(v^{†}Av{)}^{\ast }=v^{†}{A}^{†}v=-v^{†}Av$. Consequently $\overline{\mu }=-\mu$, proving that every eigenvalue of $A$ is purely imaginary, with zero permitted.