$\left\| \begin{pmatrix} 0 &A\\ A^* &0 \end{pmatrix}\right\|=\|A\|.$

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

Attempt \begin{align*} \left\| \begin{pmatrix} 0 &A\\ A^* &0 \end{pmatrix}\right\| &= \sup_{\substack{a\in E^2\\\|a\|^2=1}} \left\| \begin{pmatrix} 0 &A\\ A^* &0 \end{pmatrix}\cdot a\right\|\\ &= \sup_{\substack{x\in E,y\in E\\\|x\|^2+\|y\|^2=1}} \left\| \begin{pmatrix} 0 &A\\ A^* &0 \end{pmatrix}\cdot\begin{pmatrix} x\\y \end{pmatrix}\right\|\\ &= \sup_{\substack{x\in E,y\in E\\\|x\|^2+\|y\|^2=1}} \left\| \begin{pmatrix} Ay\\A^*x \end{pmatrix}\right\|\\ &= \sup_{\substack{x\in E,y\in E\\\|x\|^2+\|y\|^2=1}} \|Ay\|^2+\|A^*x\|^2=1\\ &=????\|A\| \end{align*}

Answer 1

Let $B$ be the big matrix, then $\|B \binom{x}{y} \|^2 = \|Ay\|^2+\|A^*x\|^2 = \|Ay\|^2+\|Ax\|^2 \le \|A\|^2 ( \|x\|^2 + \|y\|^2)$.

It is immediate from this that $\|B\| \le \|A\|$, and from $\|B \binom{0}{y} \|^2 = \|Ay\|^2$ that $\|B\| = \|A\|$.

Addendum:

Note that $\|A^*x\| = \sup_{\|u\| \le 1} | \langle u, A^* x \rangle = \sup_{\|u\| \le 1} | \langle Au, x \rangle \le \|A\| \|x\|$, so $\|A^*\| \le \|A\|$. Since $(A^*)^* = A$, we get $\|A\| \le \|A^*\|$.

Answer 2

An other way to see this: $$\begin{pmatrix} 0 & A\\ A^* & 0\end{pmatrix}$$ corresponds to the operator $X$ on $E\oplus E$ given by $$X(\xi\oplus \eta) = A\eta \oplus A^*\xi$$ i.e., $X$ is the composition of the switch map $\xi\oplus \eta \mapsto \eta \oplus \xi$ (which is a unitary!) with the direct sum operator $A\oplus A^*$. Hence, the norm of your matrix is simply $\|A \oplus A^*\| = \max\{\|A\|, \|A^*\|\}= \|A\|$.