Eigenvector of skew-hermitian matrix 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
$$
\sum_r \bigg| \sum_{s\neq r} \frac{\bar{z}_s}{\lambda_r-\lambda_s} \bigg|^2 \leq \frac{\pi^2}{\delta^2} \sum_r |z_r |^2.
$$
 A: Iwaniec and Kowalski’s use of the term "extremal vector" here is somewhat unfortunate in that this term already has a standard and quite different meaning in the Perron Frobenius theory for nonnegative matrices.   However, we can see that this context does not apply to the statement you are contemplating as the matrix $A := (\mu_{r,s})$, while irreducible, is clearly not nonnegative under the conditions of Iwaniec and Kowalski’s Lemma 7.8 (i.e., $\lambda_i$ distinct real numbers).
If we look at the left-hand side of the $(*)$ inequality you’ve written down, we see that the norm that Iwaniec and Kowalski are referring to is the spectral matrix norm.  So, we can infer that their usage of the term “extremal vector” is informal and refers to the identification of a vector $x\ne 0$ for which the induced matrix norm achieves an equality: namely, a vector $x$ such that $\left \| A \right \| _2 = \sup \limits _{x \ne 0} \frac{\left \| A x\right \| _2}{\left \| x\right \| _2}$.  In this case, since the matrix $A$ is skew-Hermitian (in fact, real skew-symmetric), we know that $A$ is normal and so we can maximize the left-hand side of $(*)$ by selecting an eigenvector of $A$ corresponding to the largest eigenvalue of $A$ in magnitude.  
