# Matrix representation of the skew-symmetric operator

Assume that $$A$$ is a skew-symmetric (skew-hermitian) operator on the finite dimensional unitary vector space $$V$$. I'm interested in the matrix representation of this operator. I found that there exists an orthonormal basis such that entries on the principal diagonal in the matrix are all equal to $$0$$ and all $$2 \times 2$$ blocks on diagonal are of the form $$\left( \begin{matrix} 0 & -a\\ a &0 \end{matrix} \right)$$.

My question is: does there exist a basis in which the matrix of the operator is a diagonal matrix of the form $$\left( \begin{matrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0& \lambda_n \end{matrix} \right)$$ where $$\lambda_i$$ are eigenvalues?

Yes, such a basis always exists since $$A$$ is normal if it is skew-symmetric. Proof: $$A^* = -A \iff A^*A = -AA = A(-A) = AA^*$$ And by the spectral theorem, normal matrices are unitarily diagonalizable which can be interpreted as - there always exists a basis in which the transformation matrix corresponding to the transformation that $$A$$ induces is diagonal.