Why $\omega\left[\begin{pmatrix} 0 &Q\\ R &0 \end{pmatrix}\right]\leq\omega\left[\begin{pmatrix} P &Q\\ R &S \end{pmatrix}\right]$? 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 $E$. The numerical radius of an operator $T\in\mathcal{L}(E)$ is given by
$$\omega(T)=\sup_{\|x\|=1}|\langle Tx, x\rangle|.$$

Let $P,Q,R,S\in\mathcal{L}(E)$. I want to prove that
$\omega\left[\begin{pmatrix}
0 &Q\\
R &0
\end{pmatrix}\right]\leq\omega\left[\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}\right].$
Here $\begin{pmatrix}
0 &Q\\
R &0
\end{pmatrix},\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}\in \mathcal{L}(E\oplus E)$.

My attempt: One can remark that
$$
 \begin{pmatrix}
0 &Q\\
R &0
\end{pmatrix} = \frac{1}{2} \begin{pmatrix}
P &Q\\
R &S
\end{pmatrix} + \frac{1}{2} \begin{pmatrix}
-P &Q\\
R &-S
\end{pmatrix},
$$
This implies that
$$
\omega\left[\begin{pmatrix}
0 &Q\\
R &0
\end{pmatrix}\right] \leq \frac{1}{2} \omega\left[\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}\right] + \frac{1}{2} \omega\left[\begin{pmatrix}
-P &Q\\
R &-S
\end{pmatrix}\right].
$$
But I'm facing difficulties to prove that
$$\omega\left[\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}\right]=\omega\left[\begin{pmatrix}
-P &Q\\
R &-S
\end{pmatrix}\right].$$
Let $T=\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}$. I want to find an unitary operator $U$ such that
$$U^*TU=\begin{pmatrix}
-P &Q\\
R &-S
\end{pmatrix}.$$
In this case we get the desired result since
$$\omega(U^*TU)=\omega(T).$$

You might not be able to find the desired $U$.  E.g. if $P=S=I$, and $Q=R=0$, then
$$
  \pmatrix{P & Q \cr R & S}
  $$
is not conjugate to
$$
  \pmatrix{-P & Q \cr R & -S},
  $$
since they have different eigenvalues.  However, letting
$$
  V=\pmatrix{-I & 0 \cr 0 & I},
  $$
one has
$$
  V\pmatrix{P & Q \cr R & S}V^{-I} =   \pmatrix{P & -Q \cr -R & S} = -\pmatrix{-P & Q \cr R & -S}
  $$
so you get what you want because $\omega (-T)=\omega (T)$.
