If $A$ is in $\mathbb R^{n \times n}$, then is $A=-A^*$ diagonalizable?
2026-03-27 04:34:18.1774586058
If $A$ is in $\mathbb R^{n \times n}$, then is $A=-A^*$ diagonalizable?
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If $A = -A^*$, then $A$ is normal: $$A A^* = A (-A) = (-A) A = A^* A .$$ In particular, $A$ is diagonalizable (in fact, unitarily diagonalizable) over $\Bbb C$.
The condition $A = -A^*$ implies that the eigenvalues of $\lambda$ are imaginary, however, so a matrix $A$ satisfying it is diagonalizable over $\Bbb R$ iff all of the eigenvalues are zero, that is, iff $A = 0$.