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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?

Thanks a lot in advance.

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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.

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  • $\begingroup$ Does it also hold for complex unitary space? And one more thing which I forgot to write in the question: does the representation with zeros and 2×2 skew-symmetric blocks on principal diagonal exists in complex vector space or only in real space? $\endgroup$
    – user121
    Commented Jun 15, 2022 at 16:10
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    $\begingroup$ @user121 It only holds in complex unitary space since (non-zero) skew-symmetric operators have non-real eigenvalues $\endgroup$
    – Ben Grossmann
    Commented Jun 15, 2022 at 16:27

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