A self adjoint matrix of operators

Question

Let $F$ be a complex Hilbert space. We recall that an operator $A\in\mathcal{B}(F)$ is said to be hyponormal if $A^*A\geq AA^*$ (i.e. $\langle (A^*A-AA^*)z,z \rangle\geq 0$ for all $z\in F$).

A pair $S=(S_1,S_2)\in\mathcal{B}(F)^2$ is called hyponormal if $$S'=\begin{pmatrix}[S_1^*, S_1] & [S_2^*,S_1]\\ [S_1^*, S_2 ]& [S_2^*, S_2] \end{pmatrix}$$ is positive on $F\oplus F$ (i.e. $\langle S'x,x \rangle\geq 0$ for all $x\in F\oplus F$.

How to show that $S'$ is self-adjoint?

Thank you.

Answer 1

Any positive operator $T$ is selfadjoint: $$ \langle T^*x,x\rangle=\langle x,Tx\rangle=\langle Tx,x\rangle, $$ where the last equality is due to $T$ being positive. Now, by polarization, $\langle T^*x,y\rangle=\langle Tx,y\rangle$ for all $x,y$, so $T^*=T$.