Let that $T \in \mathcal{B}(H)$ where H is a Hilbert space, and on $H \oplus H$ . Show that $B$ is self adjoint. Let $T \in\mathcal{B}(H)$ where $H$ is a Hilbert space, and on $H \oplus H$ we define the next operator:
$$B= \begin{bmatrix}0&iT\\−iT^∗&0\end{bmatrix}.$$
Prove that $\|B\|=\|T\|$ and that $B$ is self adjoint.
I have to use the operator norm for $B$ and $T$, also the inner product defined by $H \oplus H$ that I am using is: $$\langle(x_1,x_2),(y_1,y_2)\rangle = \langle x_1,y_1\rangle_{H} + \langle x_2,y_2 \rangle_{H}.$$
NOTE: I figured out that to prove any of the arguments that I'm asked for, first I have to evaluate $B$ in an element of $T$, and due to the space where I'm working, I am left with something like this:
$$B[(u,v)]=(iTv,-iT^*u)$$
after that I started working with the definition of norm operator to try to reach the equality I'm asked for but, I do not see a straight path to reach $\|T\|$ as I'm left with the norm of $Tv$ and $Tu$ separated. And when I try to prove that $B$ is self-adjoint I'm left with this:
\begin{align*}\langle B(u,v),(u,v)\rangle & =\langle (iTv,-iT^*u),(u,v)\rangle =\langle iTv,u \rangle +\langle -iT^*u,v \rangle \\ & =\langle v,-iT^*u \rangle +\langle u,iTv\rangle =\langle (v,u),(iT^*u,-iTv)\rangle.\end{align*}
I hoped that the last equality would give me $B$ again but I am left with something similar I do not know if this was a good idea to prove the self-adjointness of $B$. I am literally super stuck, If anyone can help me I will appreciate your help.
 A: Here is why $B$ is self-adjoint: you've already shown that
$$
\langle B(u,v),(u,v)\rangle = \langle (v,-iT^*u)\rangle +
\langle (u,iTv)\rangle.
$$
Changing the order of summation,
$$
\langle B(u,v),(u,v)\rangle = \langle (u,iTv)\rangle +\langle (v,-iT^*u)\rangle = \langle (u,v),B(u,v)\rangle,
$$
so $\langle B(u,v),(u,v)\rangle \in \mathbb{R}$, and, by polarization identity, $B$ is symmetric.
Since $T$ is bounded, both $T$ and $T^*$ must be everywhere defined (i.e. $D(T)=D(T^*)=H$). As a result, $B$ is again everywhere defined, and so $D(B)=D(B^*)=H \oplus H$.
Since $B$ is symmetric, and $D(B)=D(B^*)$, $B$ is self-adjoint.
A: $$
\begin{align}
\|B\|^2 &\stackrel{\text{def}}{=} \sup_{\|(u,v)\|_{H\oplus H}=1}\langle B(u,v),B(u,v)\rangle_{H\oplus H} \\
&= \sup_{\|u\|_H^2+\|v\|_H^2=1}\langle (iTv,-iT^*u),(iTv,-iT^*u)\rangle_{H\oplus H} \\
& = \sup_{\|u\|_H^2+\|v\|_H^2=1}(\|Tv\|_H^2 + \|T^*u\|_H^2)\\
&= \sup_\varphi \mathop{\sup_{\|u\|_H=\cos\varphi}}_{\|v\|_H=\sin\varphi} (\|Tv\|_H^2 + \|T^*u\|_H^2) \\
& = \sup_\varphi \left(\sup_{\|v\|_H=\sin\varphi}\|Tv\|_H^2 + \sup_{\|u\|_H=\cos\varphi}\|T^*u\|_H^2 \right) \\[3mm]
&=\sup_\varphi \left(\|T\|^2\sin^2\varphi + \|T^*\|^2\cos^2\varphi \right).
\end{align}
$$
Since $T$ is bounded, $\|T^*\| = \|T\|$.
