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

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

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

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• 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.) – Minus One-Twelfth Mar 2 at 20:41
• @MinusOne-Twelfth Surely not over the reals, unless $A=0$. – egreg Mar 2 at 22:34

## 1 Answer

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 $$A = 0$$.