map is an isometry we get $$\|A\| \le \sup \left [ \sum_{i=1}^n \left (\sum_{j=1}^n \|a_{ij}\|
\|x_j\|_\mathcal{H}\right )^2 \right ]^\frac{1}{2}\,,$$ or, equivalently, $$\|A\|\le \sup_{(\xi_1,\ldots ,\xi_n)\in \Xi^n} \left [ \sum_{i=1}^n \left (\sum_{j=1}^n \|a_{ij}\|
\xi_j \right )^2 \right ]^\frac{1}{2}\,,$$ where $\Xi^n:= \{(\xi_1, \ldots , \xi_n) \,|\, \xi_i \in \mathbb{R}_{+} \,\, \forall i
\,\,\, \textnormal{and} \,\,\, \sum_{i=1}^n \xi_i^2 = 1 \}$. On the other hand, $$\|\hat A\|_2 = 
\sup_{(\xi_1,\ldots ,\xi_n)\in S^n} \left [ \sum_{i=1}^n \left |\sum_{j=1}^n \|a_{ij}\|
\xi_j \right |^2 \right ]^\frac{1}{2}\,,$$ where $S^n$ denotes the unit sphere in $\mathbb{C}^n$. Observing that $\Xi^n \subset S^n$, we conclude that $\|A\| \le \|\hat A\|_2$.

To prove the second item we use the characterizations $\rho (A) = \lim_{m\to \infty}\|A^m\|^{\frac{1}{m}}$, $\rho (\hat{A}) = \lim_{m\to \infty}\|\hat{A}^m\|_2^{\frac{1}{m}}$ and the fact that $\|A^m\|\le \|\widehat{A^m}\|_2$, which we just proved. A simple inductive argument shows that $\|\widehat{A^m}\|_2 \le \|\hat{A}^m\|_2$ for all $m = 1, 2, \ldots $, thus yielding the desired result.

A version of item 1 of the previous theorem was proved by A. Ostrowski in [@ostrowski], in the context of matrices of linear operators on normed vector spaces. Related results can also be found in [@Gil] and [@Tretter].

\[remark2\] If $A$ is Hermitian or (shifted) skew-Hermitian, then $\hat A$ is real 