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Let
$$T = beginbmatrix
T_11 & T_12\
T_21 & T_22
endbmatrix,quad
S = beginbmatrix
S_11 & S_12\
S_21 & S_22
endbmatrix$$
be two positive operators on $Eoplus E$, where $E$ is a complex Hilbert space.

Note that the inner product on $Eoplus E$ is defined as follows: If $x=beginpmatrix x_1\ x_2endpmatrixin Eoplus E$ with $x_1,x_2in E$, and $x'=beginpmatrixx_1'\ x_2'endpmatrix$ similarly, then
$$langle x,x'rangle_Eoplus E:= langle x_1,x_1'rangle_E +langle x_2,x_2'rangle_E.$$

I want to prove that
$$U = beginbmatrix
T_11otimes S_11 & T_12otimes S_12\
T_21otimes S_21 & T_22otimes S_22
endbmatrix$$
is also positive where $T_ijotimes S_ij$ 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 $Eoplus Esimeq Eotimesmathbb C^2$.

Lemma. Let $H$ be any complex Hilbert space (so, e.g., $H=Eotimes mathbb C^2$ which recovers our case). If $T,Sinmathcal B(H)$ are positive semi-definite, then the same holds for $Totimes S inmathcal B(Hotimes H),$.

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

Now of course $Uneq Totimes S$ but $colorblueU$ is actually "contained" within $Totimes S$ due to
$$
Totimes S=beginpmatrix
colorblueT_11otimes S_11 & T_11otimes S_12 &T_12otimes S_11 &colorblueT_12otimes S_12 \
T_11otimes S_21 & T_11otimes S_22 &T_12otimes S_21 &T_12otimes S_22 \
T_21otimes S_11 & T_21otimes S_12 &T_22otimes S_11&T_22otimes S_12\
colorblueT_21otimes S_21 & T_21otimes S_22 &T_22otimes S_21 &colorblueT_22otimes S_22endpmatrix
$$
so we only have to "extract the blocks we care about", loosely speaking. For this, consider
$$
P=beginpmatrixoperatornameid_Eotimes E&0\0&0\0&0\0&operatornameid_Eotimes Eendpmatrixinmathcal B(Eotimes Eotimes mathbb C^2,Eotimes Eotimesmathbb C^4),.
$$
so $P$ embeds $Eotimes Eotimes mathbb C^2$ into $Eotimes Eotimesmathbb C^4$ via $(x,y)mapsto P(x,y)=(x,0,0,y)$. With this $$boxedU=P^dagger (Totimes S)P$$ 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 $Ainmathcal B(H)$ $Binmathcal B(G,H)$. If $Ageq 0$, then $B^dagger AB geq 0$.

Proof. By assumption, $A=tilde A^daggertilde A$ for some $tilde Ainmathcal B(H)$ so $B^dagger AB=B^dagger(tilde A^daggertilde A)B=(tilde AB)^dagger tilde ABgeq 0$. $square$

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