= Solution
Use the <complex inner product> $\langle a,b\rangle=\sum_j\overline{a_j}b_j=a^\dagger b$, conjugate-linear in its first argument. For $a\ne0$, put $c=\langle a,b\rangle/\langle a,a\rangle$. Positivity gives
$$
0\leq\|b-ca\|^2=\|b\|^2-\frac{|\langle a,b\rangle|^2}{\|a\|^2}.
$$
Hence the <Cauchy-Schwarz inequality> is $|\langle a,b\rangle|\leq\|a\|\|b\|$. If $a=0$ it is immediate; equality holds precisely when the two <vectors> are <linearly dependent>.
For a <unitary matrix>, $U^\dagger U=I$, so $\|Ua\|^2=a^\dagger U^\dagger Ua=\|a\|^2$. If $Uv=\lambda v$ with a nonzero <eigenvector>, this implies $|\lambda|\|v\|=\|v\|$. Thus every <eigenvalue> satisfies
$$
\boxed{|\lambda|=1.}
$$
For the one-parameter family, expand unitarity to first order: $(I+tA^\dagger+O(t^2))(I+tA+O(t^2))=I$, so $A^\dagger+A=0$. Thus $A$ is a <skew-Hermitian matrix>, an element of the <unitary Lie algebra>. If $Av=\mu v$, then $v^\dagger Av=\mu\|v\|^2$, whereas complex conjugation gives $(v^\dagger Av)^*=v^\dagger A^\dagger v=-v^\dagger Av$. Consequently $\overline\mu=-\mu$, proving that every <eigenvalue> of $A$ is purely imaginary, with zero permitted.
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