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 $E\otimes E$.

It is not difficult to show that $T\otimes S\in \mathcal{L}(E\overline{\otimes}E)$ is hyponormal if and only if $T\in\mathcal{L}(E)$ is hyponormal and $S\in\mathcal{L}(E)$ is hyponormal.

Recall that a 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 $A\otimes B:=(A_1\otimes B_1,A_2\otimes B_2)$. Is the following equivalence true? $$A\otimes B\;\text{is hyponormal if and only if}\;A\in\mathcal{L}(E)^2\;\text{ is hyponormal and}\;B\in\mathcal{L}(E)^2\;\text{ is hyponormal}. $$