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 $S=(S_1,S_2)$ hyponormal, why $S_1$ and $S_2$ are hyponormal?
Thank you.
You want to verify that $\langle (S_1^* S_1-S_1S_1^*) x, x\rangle≥0$ as well as $\langle (S_2^* S_2-S_2S_2^*) x, x\rangle≥0$ follows for any $x\in F$. Well for any $x\in F$ look at the vector $\begin{pmatrix}x\\0\end{pmatrix}$ in $F\oplus F$. You have $S'\begin{pmatrix}x\\0\end{pmatrix} = \begin{pmatrix} [S_1^*,S_1]x\\ [S_1^*,S_2]x\end{pmatrix}$. Taking the scalar product with $\begin{pmatrix} x\\0\end{pmatrix}$ recovers $$\langle [S_1^*,S_1]x,x\rangle_F=\langle S'\begin{pmatrix}x\\0\end{pmatrix},\begin{pmatrix}x\\0\end{pmatrix}\rangle_{F\oplus F} ≥0.$$ Thus $[S_1^*,S_2]$ is positive and $S_1$ is hyponormal. Do the same with $\begin{pmatrix}0\\x\end{pmatrix}$ to recover that $S_2$ is hyponormal.