The tensor product of two blocks of positive operators is positive

functional-analysisoperator-theorytensor-products

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 $S_{ij}$.

Best Answer

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.

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