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.
Best Answer
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.