0
$\begingroup$

Sorry if this is a duplicate, but I tried searching on this but only found results dealing with skew symmetric real matrices. So $A\in\mathbb{C}^{n\times n}$ is called skew-Hermitian if $A^*=-A$. I'm trying to formulate a spectral theorem for these kind of matrices. It is not difficult to see $A$ is normal, thus there exists a $U\in O_n(\mathbb{C})$ such that $U^*AU=\Lambda$, where $\Lambda$ is a diagonal matrix with the eigenvalues of $A$ as its diagonal elements. Also, by quickly writing out, with $(\lambda,v)$ a pair of an eigenvalue and corresponding eigenvector, we see that the inner product $\langle A^*v,v\rangle=\langle v,Av\rangle$ implies $-\overline\lambda=\lambda$ and thus $\mathrm{Re}(\lambda)=0$.

It is well known that if $A\in\mathbb{R}^{n\times n}$ is skew symmetric, that all eigenvalues come in pairs, and proofs can be found on this site. However, is this also true for $A\in\mathbb{C}^{n\times n}$ such that $A^*=-A$? Thus is $\lambda$ is an eigenvalue, is $-\lambda$ also an eigenvalue? And if so, how to prove this? I couldn't think of a proof. I'm looking for a hint, proof or a counterexample.

$\endgroup$
  • 1
    $\begingroup$ $A$ is skew-Hermitian iff $iA$ is Hermitian. $\endgroup$ – Lord Shark the Unknown Apr 20 '18 at 11:59
  • $\begingroup$ Thanks you! I was aware of that fact, and that would be sufficient to prove the spectral theorem I formulated. However, does it also help to see that the eigenvalues do not (necessarily) come in pairs? Obviously for a Hermitian matrix, the eigenvalues are real and do not (necessarily) come in pairs, but is it true that the eigenvalues of $A$ are the same as those of $iA$? $\endgroup$ – Václav Mordvinov Apr 20 '18 at 12:07
  • 1
    $\begingroup$ The eigenvalues of $iA$ are $i$ times the eigenvalues of $A$. $\endgroup$ – Lord Shark the Unknown Apr 20 '18 at 12:44
2
$\begingroup$

A skew symmetric matrix (over any field) is a matrix that satisfies $A^\top=-A$. (If the field has characteristic two, some people also require that $A$ has a zero diagonal.) A complex matrix $A$ that satisfies $A^\ast=-A$ is known not as a skew-symmetric matrix, but as a skew-Hermitian matrix.

The eigenvalues of a skew-Hermitian matrix do not necessarily occur in pairs. E.g. in the scalar (i.e. $1\times1$) case, $i$ is skew-Hermitian and it is the only eigenvalue of itself. For higher dimensions, consider $\operatorname{diag}(i,2i,\ldots,ni)$ for instance.

$\endgroup$
  • $\begingroup$ Thank you! I will make an edit to clearify this to other readers. I was searching to long for a proof because I assumed the eigenvalues had to come in pairs. $\endgroup$ – Václav Mordvinov Apr 20 '18 at 11:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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