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 Keister Mar 4 at 14:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Servaes, Cesareo, YiFan, Alex Provost, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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

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 $A = 0$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.