Corollary 3

Let E be a finite-dimensional complex vector space, $\phi$ a non-degenerate Hermitian sesquilinear form on E and U the unitary group of $\phi$. Then U is a real Lie subgroup of $\mathbf{GL}(E)$ and $L(U)$ is the set of $x \in \mathscr{L}(E)$ such that $ix$ is Hermitian.

When $E \neq \{0\}$, U is not a Lie subgroup of the complex Lie group $\mathbf{GL}(E)$, for $L(U)$ is not a complex vector subspace of $\mathscr{L}(E)$.