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 $x\in F\oplus F$.
If $S_1$ and $S_2$ are hyponormal. Is $S=(S_1,S_2)$ hyponormal?
Thank you.
Assume $S_1, S_2$ are self-adjoint but not necessarily commuting. Self-adjoint operators are normal and thus hyponormal. We find $$S'=\begin{pmatrix} 0& S_2S_1-S_1S_2\\ S_1S_2-S_2S_1&0\end{pmatrix}$$ and with $x=x_1\oplus x_2$:
\begin{align} \langle S'x,x\rangle &= \langle S_2S_1 x_2,x_1\rangle -\langle S_2S_1 x_1,x_2\rangle + \langle S_1S_2 x_1,x_2\rangle -\langle S_1S_2 x_2,x_1\rangle\\ &=2\mathrm{Re}(\langle S_1 x_2,S_2x_1\rangle) -2\mathrm{Re}(\langle S_1x_1,S_2 x_2\rangle). \end{align}
This allows us to construct a counter-example. Let $F=\Bbb C^2$ and $$S_1 = \begin{pmatrix} 0&1\\1&0\end{pmatrix}, \qquad S_2=\begin{pmatrix}0&0\\0&1\end{pmatrix}, \qquad x_1=\begin{pmatrix}1\\0\end{pmatrix}, x_2=\begin{pmatrix}0\\1\end{pmatrix}.$$ Then $S_1 x_1 = x_2, S_1x_2=x_1$ and $S_2x_1=0, S_2x_2=x_2$. Thus $\langle S'x,x\rangle = 2\cdot 0 -2\cdot \langle x_2,x_2\rangle = -2$, which is certainly not $≥0$.