 Solution

Anti-self-adjoint means $T^{\ast }=-T$. Unitary means $T^{\ast }T=T{T}^{\ast }=I$, so $T^{-1}=T^{\ast }$. For a unitary operator $\Vert Tu{\Vert }^2=⟨u,T^{\ast }Tu⟩=\Vert u{\Vert }^2$, and the space is nonzero. Thus $\Vert T\Vert =1$.

The series for the exponential converges absolutely in the Banach algebra $B(V)$, since ${\sum }_n\Vert T^n/n!\Vert \le e^{\Vert T\Vert }$. The adjoint is norm continuous and conjugate linear, and its action on these real scalar coefficients gives $(e^T{)}^{\ast }=e^{T^{\ast }}$. The absolutely convergent Cauchy product givesbecause its degree-$n$ coefficient for $n>0$ is $T^n{\sum }_{j=0}^n(-1{)}^j/[j!(n-j)!]=0$. If $T^{\ast }=-T$, these identities prove $e^T$ is unitary.

If $T$ is invertible, zero is outside the spectrum of $T^{-1}$. For $\mu \ne 0$ the factorizationshows that the left side is invertible exactly when the final factor is. HenceFor unitary $T$, both $T$ and $T^{-1}=T^{\ast }$ have norm one. The norm bound on each spectrum gives $|\lambda |\le 1$ and $|{\lambda }^{-1}|\le 1$, forcing $\sigma (T)$ to lie on the unit circle.

The converse fails: on ${\mathbb{C}}^2$ with its usual inner product takeIts spectrum is $\{1\}$, but $T^{\ast }T=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)\ne I$, so it is not unitary.