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If $A$ is in $\mathbb R^{n \times n}$, then is $A=-A^*$ diagonalizable?

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closed as off-topic by Servaes, Cesareo, YiFan, Alex Provost, Adrian Keister Mar 4 at 14:27

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  • $\begingroup$ Not necessarily over the reals. For example, try and see what happens if $A = \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$. (I'm assuming $A^*$ is the adjoint, which is just the transpose in the real case.) $\endgroup$ – Minus One-Twelfth Mar 2 at 20:41
  • $\begingroup$ @MinusOne-Twelfth Surely not over the reals, unless $A=0$. $\endgroup$ – egreg Mar 2 at 22:34
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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 $A = 0$.

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