# The tensor product of two blocks of positive operators is positive

Let $$T = \begin{bmatrix} T_{11} & T_{12}\\ T_{21} & T_{22} \end{bmatrix},\quad S = \begin{bmatrix} S_{11} & S_{12}\\ S_{21} & S_{22} \end{bmatrix}$$ be two positive operators on $$E\oplus E$$, where $$E$$ is a complex Hilbert space.

Note that the inner product on $$E\oplus E$$ is defined as follows: If $$x=\begin{pmatrix} x_1\\ x_2\end{pmatrix}\in E\oplus E$$ with $$x_1,x_2\in E$$, and $$x'=\begin{pmatrix}x_1'\\ x_2'\end{pmatrix}$$ similarly, then $$\langle x,x'\rangle_{E\oplus E}:= \langle x_1,x_1'\rangle_E +\langle x_2,x_2'\rangle_E.$$

I want to prove that $$U = \begin{bmatrix} T_{11}\otimes S_{11} & T_{12}\otimes S_{12}\\ T_{21}\otimes S_{21} & T_{22}\otimes S_{22} \end{bmatrix}$$ is also positive where $$T_{ij}\otimes S_{ij}$$ denotes the tensor product of the operators $$T_{ij}$$ and $$S_{ij}$$.

This might not be the easiest way to prove this but it is the best I could come up with after 10 minutes of thinking about the problem. There are two steps to this idea. For this, be aware that $$E\oplus E\simeq E\otimes\mathbb C^2$$.

Lemma. Let $$H$$ be any complex Hilbert space (so, e.g., $$H=E\otimes \mathbb C^2$$ which recovers our case). If $$T,S\in\mathcal B(H)$$ are positive semi-definite, then the same holds for $$T\otimes S \in\mathcal B(H\otimes H)\,$$.

Proof. Use that an element $$X$$ in $$\mathcal B(H)$$ is positive (semi-definite)--denoted by $$X\geq 0$$--if and only if $$X=Y^\dagger Y$$ for some $$Y\in \mathcal B(H)$$. Due to $$T,S\geq 0$$ we know $$T=\tilde T^\dagger\tilde T$$, $$S=\tilde S^\dagger\tilde S$$ which implies $$T\otimes S=( \tilde T^\dagger\tilde T \otimes \tilde S^\dagger\tilde S)=(\tilde T\otimes\tilde S)^\dagger (\tilde T\otimes\tilde S)\geq 0$$ because it is also of positive form. $$\square$$

Now of course $$U\neq T\otimes S$$ but $$\color{blue}U$$ is actually "contained" within $$T\otimes S$$ due to $$T\otimes S=\begin{pmatrix} {\color{blue}{T_{11}\otimes S_{11}}} & T_{11}\otimes S_{12} &T_{12}\otimes S_{11} &{\color{blue}{T_{12}\otimes S_{12}}} \\ T_{11}\otimes S_{21} & T_{11}\otimes S_{22} &T_{12}\otimes S_{21} &T_{12}\otimes S_{22} \\ T_{21}\otimes S_{11} & T_{21}\otimes S_{12} &T_{22}\otimes S_{11}&T_{22}\otimes S_{12}\\ {\color{blue}{T_{21}\otimes S_{21}}} & T_{21}\otimes S_{22} &T_{22}\otimes S_{21} &{\color{blue}{T_{22}\otimes S_{22}}}\end{pmatrix}$$ so we only have to "extract the blocks we care about", loosely speaking. For this, consider $$P=\begin{pmatrix}\operatorname{id}_{E\otimes E}&0\\0&0\\0&0\\0&\operatorname{id}_{E\otimes E}\end{pmatrix}\in\mathcal B(E\otimes E\otimes \mathbb C^2,E\otimes E\otimes\mathbb C^4)\,.$$ so $$P$$ embeds $$E\otimes E\otimes \mathbb C^2$$ into $$E\otimes E\otimes\mathbb C^4$$ via $$(x,y)\mapsto P(x,y)=(x,0,0,y)$$. With this $$\boxed{U=P^\dagger (T\otimes S)P}$$ which is readily verified by, e.g., multiplying out the corresponding "matrices". With this we are only one step away from our desired result.

Lemma. Let $$G,H$$ be complex Hilbert spaces and $$A\in\mathcal B(H)$$ $$B\in\mathcal B(G,H)$$. If $$A\geq 0$$, then $$B^\dagger AB \geq 0$$.

Proof. By assumption, $$A=\tilde A^\dagger\tilde A$$ for some $$\tilde A\in\mathcal B(H)$$ so $$B^\dagger AB=B^\dagger(\tilde A^\dagger\tilde A)B=(\tilde AB)^\dagger \tilde AB\geq 0$$. $$\square$$

Finally, we know that $$T\otimes S\geq 0$$ (first lemma) so $$0\leq U=P^\dagger (T\otimes S) P\in\mathcal B(E\otimes E\otimes\mathbb C^2)=\mathcal B(E\otimes E)\otimes \mathbb C^{2\times 2}$$ which concludes the overall proof. This also "recovers" the result that principal submatrices of positive matrices are positive.

• Thank you for your answer. However I don't understand the idea of your proof. Please see page 13 of this paper: ajmaa.org/searchroot/files/pdf/v13n1/v13i1p7.pdf Mar 23 '19 at 11:28
• I checked page 13 in the manuscript you linked but I'm not sure what I'm supposed to find there. It is even shown there that if $T,S$ are positive then so is $T\otimes S$ (which is what I called Lemma 1). The problem is that $T\otimes S$ is "too large", you want positivity if you only consider the corner block-elements (which you called $U$)--and those are extracted by the embedding $P$ + using that every map of the form $X\mapsto Y^\dagger XY$ preserves positivity. If there's any concrete gap in your understanding, I'll gladly try to help you after you tell me about it. Mar 23 '19 at 12:41