I was reading a book and it says: "Since the matrix is skew-hermitian, the extremal vector is an eigenvector."
I know that skew-hermitian matrix is by definition a matrix such that its conjugate transpose is equal to its negative. And that all eigenvalues of skew-hermitian matrices are purely imaginary.
But what does it even mean that a vector is "extremal vector" and why such a vector is an eigenvector of skew-hermitian matrix?
To add more context the whole thing goes like this:
Note now that the estimate (*) amounts to finding the norm of the matrix $(\mu_{r,s})$ with $\mu_{r,s}=(\lambda_r-\lambda_s)^{-1}$ and $\mu_{r,r}=0$, thus we may assume that the vector $(z_r)$ is extremal. Since the matrix is skew-hermitian, the extremal vector is an eigenvector.
The star (*) is an inequality $$ \sum_r \bigg| \sum_{s\neq r} \frac{\bar{z}_s}{\lambda_r-\lambda_s} \bigg|^2 \leq \frac{\pi^2}{\delta^2} \sum_r |z_r |^2. $$