Diagonalization skew-hermitian matrix.

Question

Let $\mathfrak{u}(n)$ be the Lie algebra of the Lie group $U(n)$. Why if $A \in \mathfrak{u}(n)$ always exists $B \in U(n)$ such that $BAB^{-1}$ is diagonal with entries $(ia_1, \cdots, ia_n)$ with $a_i \in \mathbb{R}$.

Answer 1

You probably know that normal matrices are unitarily disgonalizable i.e. for a normal matrix $N$, there exists a unitary matrix $B$ such that $B^{-1}NB = D$ where $D$ is the diagonal matrix consisting of eigenvalues of $N$.

Now a skew-hermitian matrix is normal and eigenvalues of a skew-hermitian matrix are purely imaginary and hence the result.