= Solution
Taking the <conjugate transpose> term by term in the convergent <matrix exponential> gives
$$
U^\dagger=e^{A^\dagger}=e^{-A}.
$$
The <matrices> $A$ and $-A$ <commute>, so the <matrix exponential> product law gives $U^\dagger U=e^{-A}e^A=I$ and likewise $UU^\dagger=I$. Hence $\boxed{U\text{ is unitary}}$.
For $V$, use the inverse and product rules for the <conjugate transpose>:
$$
V^\dagger=[(I+A)^{-1}]^\dagger(I-A)^\dagger=(I-A)^{-1}(I+A).
$$
The factors $I\pm A$ <commute>, as do their inverses, since they are expressions in the same <matrix>. Consequently
$$
V^\dagger V=(I-A)^{-1}(I+A)(I-A)(I+A)^{-1}=I,
$$
and similarly $VV^\dagger=I$. Therefore $\boxed{V\text{ is unitary}}$. This rational construction is the skew-Hermitian form of the <Cayley transform of a Hermitian matrix>, obtained by writing $A=iH$ with $H$ a <Hermitian matrix>.
Back to article page
