Tensor product of two positive operators

Let $$E$$, $$F$$ be two complex Hilbert spaces and $$\mathcal{L}(E)$$ (resp. $$\mathcal{L}(F)$$) be the algebra of all bounded linear operators on $$E$$ (resp. $$F$$).

The algebraic tensor product of $$E$$ and $$F$$ is given by $$E \otimes F:=\left\{\xi=\sum_{i=1}^dv_i\otimes w_i:\;d\in \mathbb{N}^*,\;\;v_i\in E,\;\;w_i\in F \right\}.$$

In $$E \otimes F$$, we define $$\langle \xi,\eta\rangle=\sum_{i=1}^n\sum_{j=1}^m \langle x_i,z_j\rangle_1\langle y_i ,t_j\rangle_2,$$ for $$\xi=\displaystyle\sum_{i=1}^nx_i\otimes y_i\in E \otimes F$$ and $$\eta=\displaystyle\sum_{j=1}^mz_j\otimes w_j\in E \otimes F$$.

The above sesquilinear form is an inner product in $$E \otimes F$$.

It is well known that $$(E \otimes F,\langle\cdot,\cdot\rangle)$$ is not a complete space. Let $$E \widehat{\otimes} F$$ be the completion of $$E \otimes F$$ under the inner product $$\langle\cdot,\cdot\rangle$$.

If $$T\in \mathcal{L}(E)$$ and $$S\in \mathcal{L}(F)$$, then the tensor product of $$T$$ and $$S$$ is denoted $$T\otimes S$$ and defined as $$\big(T\otimes S\big)\bigg(\sum_{k=1}^d x_k\otimes y_k\bigg)=\sum_{k=1}^dTx_k \otimes Sy_k,\;\;\forall\,\sum_{k=1}^d x_k\otimes y_k\in E \otimes F,$$ which lies in $$\mathcal{L}(E \otimes F)$$. The extension of $$T\otimes S$$ over the Hilbert space $$E \widehat{\otimes} F$$, denoted by $$T \widehat{\otimes} S$$, is the tensor product of $$T$$ and $$S$$ on the tensor product space, which lies in $$\mathcal{L}(E\widehat{\otimes}F)$$.

An operator $$A\in\mathcal{L}(E)$$ is said to be positive if $$\langle Ax\mid x\rangle \geq 0$$ for any $$x\in E$$.

If $$T$$ and $$S$$ are positive operators, I want to prove $$T\otimes S$$ is positive on $$E \otimes F$$.

Let $$X=\sum_{i=1}^nx_i\otimes y_i\in E\otimes F$$. Then \begin{align*} \langle (T\otimes S)X,X\rangle & =\sum_{i=1}^n\sum_{j=1}^n \langle Tx_i\mid x_j\rangle_1\langle Sy_i\mid y_j\rangle_2. \end{align*} My goal is to prove that $$\langle (T\otimes S)X,X\rangle \geq 0,$$ for any $$X \in E\otimes F$$.

I'm not sure that it can be done directly as you want. But if you know that positive operators admit positive square roots, you know that $$T=T_0^*T_0$$, $$S=S_0^*S_0$$. Then $$T\otimes S=T_0^*T_0\otimes S_0^*S_0=(T_0\otimes S_0)^*(T_0\otimes S_0)\geq0.$$
• Thanks a lot for your answer. However, I think you have shown that $T \widehat{\otimes} S$ is a positive operator in $E\widehat{\otimes}F$ because $(T_0\otimes S_0)^*$ does not make sense in $E\otimes F$ because it is not a Hilbert space. Jan 31 '19 at 18:57
• Yes, but the restriction to $E\otimes F$ is $T\otimes S$. Jan 31 '19 at 19:00