An operator on $H\times H$, with $H$ Hilbert Let $(H, \langle \cdot,\cdot\rangle_H)$ a Hilbert complex space and consider  $H\times H$ with the inner product
$$\langle (u,v),(z,w)\rangle_{H\times H}\ =\ \langle u,z\rangle_H + \langle v,w\rangle_H.$$
Let, $A\in\mathcal{L}(H,H)$ and define the operator $B:H\times H\rightarrow H\times H$ by
$$B((u,v))\ :=\ (iA(v),\ -iA^*(u)),\quad \forall\ (u,v)\in H\times H.$$
Show that $\|B\| = \|A\|$ and $B$ is self-adjoint.

I was working in that problem, and I have already proved that $B\in\mathcal{L}(H\times H, H\times H)$ and $\|B\| \leq \|A\|$, but I have had problems in order to prove that $\|B\| \geq \|A\|$, please I need hints. Thanks in advance.
 A: So you are almost there. Just note that for every $\|v \|=1$, we have $\|(0,v)\|=1$ whence
$$
\|B\|\geq \|B(0,v)\|=\|iAv\|=\|Av\|.
$$
Taking the sup over $\|v\|=1$ yields $\|B\|\geq \|A\|$ as desired. 
To prove that $B$ is self-adjoint 
$$
((u,v),B(u',v'))=(u,iAv')+(v,-iA^*u')=((iA)^*u,v')+((-iA^*)^*v,u')
$$
$$
=(-iA^*u,v')+(iAv,u')=(B(u,v),(u',v')).
$$

Here is an outline of how this would be done in Operator Algebras, using $2\times 2$ matrices over $B(H)$. It is indeed a fundamental fact that $B(H\times H)$ can be identified with $M_2(B(H))$. Note that $H\times H$ is most often denoted by $H\oplus H$. And note that the vectors of $H$ disappear from the argument.
We can identify $B$ with the following $2\times 2$ matrix 
$$
B= \pmatrix{0&iA\\-iA^*&0}
$$
with respect to the orthogonal decomposition $H\times H=H\times \{0\}\oplus \{0\}\times H$. Therefore
$$
B^*=\pmatrix{0&(-iA^*)^*\\(iA)^*&0}=B\quad\mbox{and}\quad B^*B=B^2=\pmatrix{AA^*&0\\0&A^*A}.
$$
So $B$ is self-adjoint. Now in general, for a block diagonal operator over an orthogonal decomposition, we have
$$
\Big\|\pmatrix{S&0\\0&T}\Big\|=\max \{\|S\|,\|T\|\}.
$$
Now recall the $C^*$-norm property of the operator norm of $B(H)$: $\|A^*A\|=\|AA^*\|=\|A\|^2$. Hence
$$
\|B\|^2=\|B^*B\|=\max\{\|AA^*\|,\|A^*A\|\}=\|A\|^2\quad\Rightarrow\quad \|B\|=\|A\|.
$$
