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I know that Real Skew-Symmetric matrices are diagonalisable over Complex field, need not to be over Real field. But, I am not getting breakthrough of whether Complex Skew-Symmetric matrices are always diagonalisable over Complex filed? If anyone can help? Thanks in advance!

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Complex skew-symmetric matrices are not necessarily diagonalizable. See this post for an example of a skew-symmetric matrix that fails to be diagonalizable.

In particular: for $n = 3$, we can proceed as follows. Define $$ X = \pmatrix{\frac{1 - i}2 & 0 & \frac{1+i}{2}\\ 0 & i & 0\\\frac{1+i}{2} & 0 & \frac{1-i}{2}}, $$ which is a symmetric matrix that satisfies $X^4 = I$. Define $J$ to be the Jordan matrix $$ J = \pmatrix{0&1&0\\0&0&1\\0&0&0}. $$ We can show that $XJX^{-1}$ is skew-symmetric, but it is not diagonalizable because it is similar to $J$. In particular, we find that $$ A = XJX^{-1} = \frac 12 \pmatrix{0 & -1- i & 0\\1+ i & 0 & -1+ i\\0 & 1- i & 0} $$ is skew-symmetric and non-diagonalizable.

On the other hand, it is true that skew-Hermitian matrices are necessarily diagonalizable as a consequence of the spectral theorem for normal matrices.

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  • $\begingroup$ I am unable to find that example in this attached post by you Ben Grossmann. Can you help me plz? $\endgroup$ Commented May 6, 2021 at 19:04
  • $\begingroup$ Thank you! Thanks alot! $\endgroup$ Commented May 6, 2021 at 19:06
  • $\begingroup$ You're welcome. I've also written the matrix out in full in case you find that helpful. In fact, we can get rid of the $\frac 12$ in front to come up with another valid counterexample. $\endgroup$ Commented May 6, 2021 at 19:15
  • $\begingroup$ Thanks a ton!😊 $\endgroup$ Commented May 6, 2021 at 19:23

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