Let $E$ be a complex Hilbert space. We recall that an operator $T\in\mathcal{L}(E)$ is said to be hyponormal if $[T^\*, T]\geq 0$ (i.e. $\langle (T^\*T-TT^\*)x,x \rangle\geq 0$ for all $x\in E$). Let $E\overline{\otimes}E$ denotes the completion,
endowed with a reasonable uniform cross-norm, of the algebraic tensor product $E\otimes E$.


The pair $A=(A\_1,A\_2)\in\mathcal{L}(E)^2$ is called hyponormal if
$$\varphi(A)=\begin{pmatrix}[A\_1^\*, A\_1] & [A\_2^\*,A\_1]\\
[A\_1^\*, A\_2 ]& [A\_2^\*, A\_2]
\end{pmatrix}$$
is positive on $E\oplus E$ (i.e. $\langle \varphi(A)x,x \rangle\geq 0$ for all $x\in E\oplus E$.



> 
> For $A=(A\_1,A\_2)\in\mathcal{L}(E)^2$ and $B=(B\_1,B\_2)\in\mathcal{L}(E)^2$ we consider $A\otimes B:=(A\_1\otimes B\_1,A\_2\otimes B\_2)$. Assume that $A,B\in\mathcal{L}(E)^2$ are hyponormal. Why $A\otimes B\in\mathcal{L}(E\overline{\otimes}E)^2$ is hyponormal?
> 
> 
> 



Proving the hyponormality of $A\otimes B$
I think the ``if part'' is correct and it follows from the following fact: if the operator-matrices
$$T = \begin{bmatrix}
T\_{11} & T\_{12}\\
T\_{21} & T\_{22}
\end{bmatrix},\quad
S = \begin{bmatrix}
S\_{11} & S\_{12}\\
S\_{21} & S\_{22}
\end{bmatrix}$$
are both positive, then the matrix obtained by entry-wise tensor
$$U = \begin{bmatrix}
T\_{11}\otimes S\_{11} & T\_{12}\otimes S\_{12}\\
T\_{21}\otimes S\_{21} & T\_{22}\otimes S\_{22}
\end{bmatrix}$$
is also positive. This is similar to Hadamard product of two positive matrices.


On the other hand, there seems to be an issue with the ``only if part'' (even in the case of single operators): what if you take $A=(0,0)$ and $B$ is arbitrary?


