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 $E\otimes E$.

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 $A\otimes B:=(A_1\otimes B_1,A_2\otimes B_2)$. Assume that $A,B\in\mathcal{L}(E)^2$ are hyponormal. It is true that $A\otimes B\in\mathcal{L}(E\overline{\otimes}E)^2$ is hyponormal?

Note 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.

In fact: Assume that $T$ and $S$ are hyponormal. This implies that $$(T\otimes S)^*(T\otimes S)=(T^*T)\otimes (S^*S)\geq (T\otimes S)(T\otimes S)^*.$$ Conversely we use the following result discovered by STOCHEL

Lemma: Let $A_1, A_2\in \mathcal{L}(E)$ and $B_1, B_2\in \mathcal{L}(E)$ be nonnegative operators. If $A_1\not=0$ and $B_1\not=0 $, then the following conditions are equivalent:

• $A_1\otimes B_1\leq A_2\otimes B_2$.

• There exists $c > 0$ such that $A_1 \leq cA_2$ and $B_1\leq c^{-1}B_2.$

Clearly $T\otimes S$ is hyponormal if and only if $T^*T\otimes S^*S\geq TT^*\otimes SS^*$. So by applying STOCHEL result, there exists $c>0$ such that $$cT^*T\geq TT^*\;\;\text{and}\;\;c^{-1}S^*S\geq SS^*.$$ A simple computation shows that $c=1$.

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?

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 $E\otimes E$.

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 $A\otimes B:=(A_1\otimes B_1,A_2\otimes B_2)$. I want to show or to find suitable conditions under which the following equivalence holds: $$A\otimes B\in\mathcal{L}(E\overline{\otimes}E)^2\;\text{is hyponormal}\Longleftrightarrow\;A,B\in\mathcal{L}(E)^2\;\;\text{are hyponormal}. $$

Note 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.

In fact: Assume that $T$ and $S$ are hyponormal. This implies that $$(T\otimes S)^*(T\otimes S)=(T^*T)\otimes (S^*S)\geq (T\otimes S)(T\otimes S)^*.$$ Conversely we use the following result discovered by STOCHEL

Lemma: Let $A_1, A_2\in \mathcal{L}(E)$ and $B_1, B_2\in \mathcal{L}(E)$ be nonnegative operators. If $A_1\not=0$ and $B_1\not=0 $, then the following conditions are equivalent:

• $A_1\otimes B_1\leq A_2\otimes B_2$.

• There exists $c > 0$ such that $A_1 \leq cA_2$ and $B_1\leq c^{-1}B_2.$

Clearly $T\otimes S$ is hyponormal if and only if $T^*T\otimes S^*S\geq TT^*\otimes SS^*$. So by applying STOCHEL result, there exists $c>0$ such that $$cT^*T\geq TT^*\;\;\text{and}\;\;c^{-1}S^*S\geq SS^*.$$ A simple computation shows that $c=1$.

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 $E\otimes E$.

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 $A\otimes B:=(A_1\otimes B_1,A_2\otimes B_2)$. Assume that $A,B\in\mathcal{L}(E)^2$ are hyponormal. It is true that $A\otimes B\in\mathcal{L}(E\overline{\otimes}E)^2$ is hyponormal?

Note 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.

In fact: Assume that $T$ and $S$ are hyponormal. This implies that $$(T\otimes S)^*(T\otimes S)=(T^*T)\otimes (S^*S)\geq (T\otimes S)(T\otimes S)^*.$$ Conversely we use the following result discovered by STOCHEL

Lemma: Let $A_1, A_2\in \mathcal{L}(E)$ and $B_1, B_2\in \mathcal{L}(E)$ be nonnegative operators. If $A_1\not=0$ and $B_1\not=0 $, then the following conditions are equivalent:

• $A_1\otimes B_1\leq A_2\otimes B_2$.

• There exists $c > 0$ such that $A_1 \leq cA_2$ and $B_1\leq c^{-1}B_2.$

Clearly $T\otimes S$ is hyponormal if and only if $T^*T\otimes S^*S\geq TT^*\otimes SS^*$. So by applying STOCHEL result, there exists $c>0$ such that $$cT^*T\geq TT^*\;\;\text{and}\;\;c^{-1}S^*S\geq SS^*.$$ A simple computation shows that $c=1$.

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? $$A\otimes B\;\text{is hyponormal if and only if}\;A\in\mathcal{L}(E)^2\;\text{ is hyponormal and}\;B\in\mathcal{L}(E)^2\;\text{ is hyponormal}. $$

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 $E\otimes E$.

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 $A\otimes B:=(A_1\otimes B_1,A_2\otimes B_2)$. I want to show or to find suitable conditions under which the following equivalence holds: $$A\otimes B\in\mathcal{L}(E\overline{\otimes}E)^2\;\text{is hyponormal}\Longleftrightarrow\;A,B\in\mathcal{L}(E)^2\;\;\text{are hyponormal}. $$

Note 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.

In fact: Assume that $T$ and $S$ are hyponormal. This implies that $$(T\otimes S)^*(T\otimes S)=(T^*T)\otimes (S^*S)\geq (T\otimes S)(T\otimes S)^*.$$ Conversely we use the following result discovered by STOCHEL

Lemma: Let $A_1, A_2\in \mathcal{L}(E)$ and $B_1, B_2\in \mathcal{L}(E)$ be nonnegative operators. If $A_1\not=0$ and $B_1\not=0 $, then the following conditions are equivalent:

• $A_1\otimes B_1\leq A_2\otimes B_2$.

• There exists $c > 0$ such that $A_1 \leq cA_2$ and $B_1\leq c^{-1}B_2.$

Clearly $T\otimes S$ is hyponormal if and only if $T^*T\otimes S^*S\geq TT^*\otimes SS^*$. So by applying STOCHEL result, there exists $c>0$ such that $$cT^*T\geq TT^*\;\;\text{and}\;\;c^{-1}S^*S\geq SS^*.$$ A simple computation shows that $c=1$.