Proving the hyponormality of $A\otimes B$

Question

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?

Answer 1

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?