Why $\left\| \begin{pmatrix} 0 &A\\ B &0 \end{pmatrix}\right\|\geq\max\{\|A\|,\|B\|\}$?

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

It can be verified that

$$\left\| \begin{pmatrix} 0 &A\\ B &0 \end{pmatrix}\right\|\leq\max\{\|A\|,\|B\|\}.$$

I'm facing difficulties to prove the converse inequality.

I see this result in a paper

Answer 1

Let $X=\begin{pmatrix}0&A\\B&0\end{pmatrix}$. By the $C^\ast$-identity we have $$ \lVert X\rVert^2=\lVert X^\ast X\rVert=\left\lVert\begin{pmatrix}B^\ast B&0\\0&A^\ast A\end{pmatrix}\right\rVert. $$ If $\xi_n\in E$ with $\|\xi_n\|=1$ and $\|A^\ast A\xi_n\|\to\|A\|^2$, then $\|(0,\xi_n)^T\|=1$ and $$ \left\lVert X^\ast X\begin{pmatrix}0\\ \xi_n\end{pmatrix}\right\rVert=\left\lVert \begin{pmatrix}0\\ A^\ast A\xi_n\end{pmatrix}\right\rVert=\|A^\ast A\xi_n\|\to \|A\|^2. $$ Thus $\|X\|^2\geq \|A\|^2$. The inequality $\|X\|^2\geq \|B\|^2$ can of course be proven similarly. Therefore $\|X\|\geq\max\{\|A\|,\|B\|\}$.

Answer 2

We have

$\left\{\left(\matrix{0\\ Y}\right) | \;\;\lVert Y\rVert^2=1\right\}\subseteq \left\{\left(\matrix{X\\ Y}\right) | \;\;\lVert X\rVert^2+\lVert Y\rVert^2=1\right\}$

So, by taking the sup on a bigger set,

$\sup_{\lVert X\rVert^2+\lVert Y\rVert^2=1}\left\lVert \pmatrix{0 & A\\B & 0}\pmatrix{ X \\ Y}\right\rVert\geq \sup_{\lVert Y\rVert^2=1}\left\lVert\pmatrix{0 & A\\B & 0}\pmatrix{0\\ Y}\right\rVert$

However, $\pmatrix{0 & A\\B & 0}\pmatrix{0\\ Y}=\pmatrix{AY \\ 0}$

So $\sup_{\lVert Y\rVert^2=1}\left\lVert\pmatrix{0 & A\\B & 0}\pmatrix{0\\ Y}\right\rVert=\sup_{\lVert Y\rVert^2=1}\left\lVert\pmatrix{AY \\ 0}\right\rVert$

$=\sup_{\lVert Y\rVert^2=1}\left\lVert AY \right\rVert$

$=\left\lVert A \right\rVert$

Hence

$\left\lVert \pmatrix{0 & A\\B & 0}\right\rVert\geq \left\lVert A \right\rVert$

idem for B.

Answer 3

The formula holds for $B(X),$ the bounded operators on a normed space $X.$ No $C^*$-structure is needed. Let $X\times X$ be equipped with the norm $\|x\|=\|(x_1,x_2)\|=\sqrt{\|x_1\|^2+\|x_2\|^2}.$ Consider operators $T$ on $X\times X$ given by $$Tx=\begin{pmatrix} 0 &A\\ B & 0\end{pmatrix}\begin{pmatrix} x_1 \\x_2\end{pmatrix}=\begin{pmatrix} Ax_2 \\ Bx_1\end{pmatrix}$$ where $A,B\in B(X)$ and $x_1,x_2\in X.$ Then $$ \displaylines{\|Tx\|^2=\|Ax_2\|^2+\|Bx_1\|^2 \le \|A\|^2\|x_2\|^2+\|B\|^2\|x_1\|^2 \\ \le \max\{\|A\|^2,\|B\|^2\}\,\|x\|^2}$$ Therefore $$\|T\|\le \max\{\|A\|,\|B\|\}$$

On the other hand $$T\begin{pmatrix}x_1 \\ 0\end{pmatrix}= \begin{pmatrix} 0\\Bx_1\end{pmatrix},\qquad T\begin{pmatrix}0 \\ x_2\end{pmatrix}= \begin{pmatrix} Ax_2\\0\end{pmatrix}$$ Therefore $$\|T\|\ge \|B\|, \qquad \|T\|\ge \|A\|$$ Remark The same formula is valid for any reasonable norm on the product space $X\times X,$ like for example $\ell^p$-norm.