Why $\left\| \begin{pmatrix} 0 &B\\ C &0 \end{pmatrix}\right\|\leq\left\| \begin{pmatrix} A &B\\ C &D \end{pmatrix}\right\|$?

Question

Let $E$ be a complex Hilbert space, with inner product $\langle\cdot\;, \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{L}(E)$ the algebra of all operators on $E$.

Let $A,B,C,D\in\mathcal{L}(E)$. I want to prove that $$\left\| \begin{pmatrix} 0 &B\\ C &0 \end{pmatrix}\right\|\leq\left\| \begin{pmatrix} A &B\\ C &D \end{pmatrix}\right\|.$$ Here $\begin{pmatrix} 0 &B\\ C &0 \end{pmatrix},\begin{pmatrix} A &B\\ C &D \end{pmatrix}\in \mathcal{L}(E\oplus E)$.

I read this result in this paper without citing any references.

Answer 1

$$\left\| \pmatrix{-A & B\cr C & -D\cr}\right\|= \left\| \pmatrix{A & B\cr C & D\cr}\right\|$$ because $(x,y) \mapsto (x,-y)$ is an isomorphism of $E \oplus E$, and $$ \pmatrix{0 & B\cr C & 0\cr} = \frac{1}{2} \pmatrix{A & B\cr C & D\cr} + \frac{1}{2} \pmatrix{-A & B\cr C & -D\cr}$$