Complex Skew-Symmetric Matrices, Diagonalisibility Question

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!

1 Answer

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.

• I am unable to find that example in this attached post by you Ben Grossmann. Can you help me plz? Commented May 6, 2021 at 19:04
• Thank you! Thanks alot! Commented May 6, 2021 at 19:06
• 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. Commented May 6, 2021 at 19:15
• Thanks a ton!😊 Commented May 6, 2021 at 19:23