Let $r_A$ and $r_B$ are ranks of $A$ and $B$. Using, e.g., the SVD, $A$ and $B$ can be written as
$$
A=\sum_{i=1}^{r_A}p_iq_i^*, \quad B=\sum_{j=1}^{r_B}s_jt_j^*,
$$
for some vectors $p_i$, $q_i$, $i=1,\ldots,r_A$, and $s_j$, $t_j$, $j=1,\ldots,r_B$.
We have
$$
A\circ B=\sum_{i=1}^{r_A}\sum_{j=1}^{r_B}(p_i\circ s_j)(q_i\circ t_j)^*,
$$
so $A\circ B$ is a sum of $r_Ar_B$ rank-one matrices and thus its rank cannot be higher than $r_Ar_B$.
To show that $r_Ar_B=\mathrm{rank}(A\otimes B)$, note that if $A=U_A\Sigma_AV_A^*$ and $B=U_B\Sigma_BV_B^*$ are SVDs of $A$ and $B$, respectively, then
$$
A\otimes B=(U_A\otimes U_B)(\Sigma_A\otimes \Sigma_B)(V_A\otimes V_B)^*
$$
is an SVD of $A\otimes B$. So since there are $r_A$ and $r_B$ nonzero singular values of $A$ and $B$, respectively, there are $r_Ar_B$ nonzero singular values of $A\otimes B$.
The chain of inequalities makes sense if we add in the following.
Claim: For any unit vectors $x,y$, we have
$$
\|x - y\| \geq \frac 12 \operatorname{Tr}|xx^* - yy^*|.
$$
Proof: Noting that $x$ and $y$ are unit vectors, we compute
$$
\|x - y\|^2 = (x-y)^*(x-y) = x^*x - x^*y - y^*x + y^*y \\
= 1 - \langle x, y\rangle - \langle y,x \rangle + 1 = 2(1 - \operatorname{Re}\langle x,y \rangle)
$$
So, we have $\|x - y\| = \sqrt{2}\sqrt{1 - \operatorname{Re}\langle x,y \rangle}$.
On the other hand, $A = (xx^* - yy^*)$ is a rank-2 Hermitian matrix with trace zero, which means that it has non-zero eigenvalues $\pm \lambda$ for some $\lambda>0$. We compute
$$
\begin{align*}
2\lambda^2 &= \lambda^2 + (-\lambda)^2 = \operatorname{Tr}(A^2) = \operatorname{Tr}[(xx^* - yy^*)^2]
\\ & = \operatorname{Tr}[xx^*xx^* - xx^*yy^* - yy^*xx^* + yy^*yy^*]
\\ & =
\operatorname{Tr}[x^*xx^*x - x^*yy^*x - y^*xx^*y + y^*yy^*y]
\\ &=
\langle x,x\rangle^2 - 2|\langle x,y\rangle|^2 + \langle y,y \rangle^2
\\ & =
2 - 2|\langle x,y\rangle|^2
\end{align*}
$$
We thereby conclude that $\lambda = \sqrt{1 - |\langle x,y \rangle|}$. Thus, we compute
$$
\frac 12 \operatorname{Tr}|xx^* - yy^*| = \frac 12 (2 \lambda) = \sqrt{1 - |\langle x,y \rangle|}.
$$
Thus, it suffices to show that
$$
\sqrt{2}\sqrt{1 - \operatorname{Re}\langle x,y \rangle} \geq
\sqrt{1 - |\langle x,y \rangle|}.
$$
Indeed, we have
$$
\sqrt{2}\sqrt{1 - \operatorname{Re}\langle x,y \rangle} \geq
\sqrt{1 - |\langle x,y \rangle|} \iff\\
2(1 - \operatorname{Re}\langle x,y \rangle) \geq 1 - |\langle x, y \rangle| \iff \\
1 + |\langle x, y \rangle| \geq 2 \operatorname{Re}\langle x,y \rangle \iff\\
\frac{1 + |\langle x, y \rangle|}{2} \geq \operatorname{Re}\langle x,y \rangle.
$$
The last inequality can be shown to hold as follows: by Cauchy Schwarz, $|\langle x, y \rangle| < 1$. So,
$$
\frac{1 + |\langle x, y \rangle|}{2} \geq |\langle x,y \rangle| = \sqrt{(\operatorname{Re}\langle x,y \rangle)^2 + (\operatorname{Im} \langle x,y\rangle)^2} \\
\qquad \qquad \quad\geq
\sqrt{(\operatorname{Re}\langle x,y \rangle)^2} = |\operatorname{Re}\langle x,y \rangle| \geq \operatorname{Re}\langle x,y \rangle.
$$
Best Answer
It is just a computation.
We have $$ S^\dagger=\sum_i e_i\otimes e_i^t\otimes e_i\otimes e_1^t $$ So $$ (A_1\otimes A_2)S^\dagger = \sum_i A_1e_i\otimes e_i^t\otimes A_2 e_i\otimes e_1^t $$ and \begin{align*} S(A_1\otimes A_2)S^\dagger &= \left(\sum_j e_j\otimes e_j^t\otimes e_1\otimes e_j^t\right)\left(\sum_i A_1e_i\otimes e_i^t\otimes A_2 e_i\otimes e_1^t\right)\\ &= \sum_{i,j}\left( e_j\otimes e_j^t\otimes e_1\otimes e_j^t\right)\left( A_1e_i\otimes e_i^t\otimes A_2 e_i\otimes e_1^t\right)\\ &= \sum_{i,j}\left( (e_j\otimes e_j^t)(A_1e_i\otimes e_i^t)\otimes (e_1\otimes e_j^t)(A_2 e_i\otimes e_1^t)\right)\\ &= \sum_{i,j}\left( (e_j^t A_1 e_i)(e_j\otimes e_i^t)\otimes (e_j^t A_2 e_i)(e_1 \otimes e_1^t)\right)\\ &= \sum_{i,j} (e_j^t A_1 e_i)(e_j^t A_2 e_i)\left((e_j\otimes e_i^t)\otimes (e_1 \otimes e_1^t)\right)\\ &= \left(\sum_{i,j} (e_j^t (A_1\circ A_2) e_i)(e_j\otimes e_i^t)\right)\otimes (e_1 \otimes e_1^t)\\ &= (A_1\circ A_2) \otimes (e_1 \otimes e_1^t)\\ \end{align*}