I was reading this preprint: https://arxiv.org/abs/1605.00531. The authors mention the following:
"It is known that the eigenvalues $z_k$ of any matrix $Z$ are in the rectangle $\text{Re}( z_k) \in \sigma_1$, $\text{Im}(z_k) \in \sigma_2$ where $\sigma_1$ is the range of $\frac{1}{2}(Z+Z^\dagger)$ and $\sigma_2$ is the range of $\frac{1}{2i}(Z-Z^\dagger)$."
I thought the range was a set of all linear combinations of the columns of $Z$? is $\sigma_1$ the cardinality of that set?
More generally, how can I prove their statement, or where can I find a proof for it? The reference the author used is inaccessible with my university.
Thank you very much!
The "range" here most likely refers to the numerical range of $Z$, i.e. the set of all Rayleigh quotients for $Z$.
Let $H_1=\frac12(Z+Z^\dagger)$ and $H_2=\frac{1}{2i}(Z-Z^\dagger)$. For any unit eigenvector $x$ corresponding to an eigenvalue $a+ib$ of $Z$, we have $x^\dagger Zx=a+ib$. Hence \begin{aligned} a&=\frac12\left[(a+ib)+(a+ib)^\dagger\right]=\frac12\left[x^\dagger Zx+(x^\dagger Zx)^\dagger\right]=x^\dagger H_1x,\\ b&=\frac{1}{2i}\left[(a+ib)-(a+ib)^\dagger\right]=\frac{1}{2i}\left[x^\dagger Zx-(x^\dagger Zx)^\dagger\right]=x^\dagger H_2x. \end{aligned} Therefore $a$ and $b$ lie inside the numerical ranges of $H_1$ and $H_2$ respectively.