Property of partial traces Consider the Kronecker product of $A \in M_m, B \in M_n$:
$A \otimes B = \left( \begin{matrix} a_{11}B&...&a_{1m}B\\ \vdots&\ddots\\a_{m1}B&...&a_{mm}B \end{matrix} \right)$
$A \otimes B$ can also be thought of as an $n \times n$ block matrix. This representation can be seen as $\left( \begin{matrix} T_{11}&...&T_{1n}\\ \vdots&\ddots\\T_{n1}&...&T_{nn} \end{matrix} \right)$, where each $T$ entry in the block is an $m \times m$ matrix.
If we define the two partial traces of a Kronecker product between two matrices as follows:


*

*$Tr_1(A \otimes B) = \sum_{i=1}^n T_{ii}$. We have $Tr_1 \in M_m$. 

*$Tr_2(A \otimes B) = \left( \begin{matrix} Tr(T_{11})&...&Tr(T_{1n})\\ \vdots&\ddots\\Tr(T_{n1})&...&Tr(T_{nn}) \end{matrix} \right)$. We have $Tr_2 \in M_n$.


How does one show that for any $X \geq 0$, $Im(Tr_2(X) \otimes Tr_1(X)) \supset Im(X)$ (where $X$ is Hermitian and has the form $A \otimes B$ for some $A,B$)?
 A: For clarification, following the question, let us emphasize some notations first. 
Let $M_n$ be the collection of $n\times n$ complex valued matrices, let $H_n$ be the collection of $n\times n$ complex valued Hermitian matrices, and given $X,Y\in H_n$, denote $X\ge 0\Leftrightarrow$ $X$ is positive semidefinite, and $X\ge Y\Leftrightarrow X-Y\ge 0$. Given $m,n$, for $X\in M_{mn}$, write $X=(X_{ij})_{n\times n}$, where $X_{ij}\in M_m$. Define 
$${\rm tr}_1 (X)=\sum_{i=1}^n X_{ii}\in M_m,\quad {\rm tr}_2 (X)=({\rm tr}(X_{ij}))_{n\times n}\in M_n\tag{1}$$
and
$$\mathcal P(X)={\rm tr}_2(X)\otimes {\rm tr}_1(X)\in M_{mn}.\tag{2}$$
Note that  the definitions $(1)$ and $(2)$ for $X\in M_{mn}(X)$ not only depend on the the product $mn$, but also depend on $m$ and $n$ themselves. Since there is no much ambiguity in the following, for simplicity, we will not specify the dependence on $m,n$.  

Claim: Given $X\in H_{mn}$,  $$X\ge 0\Rightarrow {\rm Im}(\mathcal P(X))\supset {\rm Im}(X).\tag{3}$$

The proof is based on the following facts. Since the argument is very long, let us omit the proof of the facts. 
Fact 1. Given  $X,Y\in H_n$ with $X,Y\ge 0$,  $${\rm Im}(X)\supset {\rm Im}(Y)\iff {\rm Ker}(X)\subset {\rm Ker}(Y)\iff X\ge c Y \quad\text{for some} \quad c>0.\tag{4}$$
Fact 2. Given $A\in H_n$ and $B\in H_m$, we have $A\otimes B\in H_{mn}$ and 
$$A\ge 0, B\ge 0 \Rightarrow A\otimes B\ge 0.\tag{5}$$
Fact 3. Given $X\in H_{mn}$, 
$$X\ge 0 \Rightarrow {\rm tr}_1(X)\ge 0,\ {\rm tr}_2(X)\ge 0.\tag{6}$$
From the definitions $(1)$, $(2)$ and the facts 2 and 3, we have
Fact 4.
Given $Y,Z\in H_n$, if $Y,Z\ge 0$, then for $X=Y+Z$,
$$\mathcal P(X)\ge \mathcal P(Y)+\mathcal P(Z).\tag{7}$$
Fact 5.(special case of Sylvester's law of inertia)
If $X\in H_n$, $X\ge 0$ and $X$ has rank $1\le r\le n$, then there exist $r$ nonzero vectors $v_1,\cdots,v_r\in\mathbb C^n$, such that 
$${\rm Im}(X)={\rm span}(v_1,\dots,v_r),\text{ and } X=\sum_{i=1}^rv_iv_i^*.\tag{8}$$
Here $v_i$ is denoted as column vector and $v_i^*$ denotes its conjugate transpose.
Proof of Claim:
Write $X=\sum_{i=1}^rv_iv_i^*$ as in $(8)$. Note that $v_iv_i^*\ge 0$. Then apply $(7)$ inductively we can obtain that
$$\mathcal P(X)\ge\sum_{i=1}^r\mathcal P(v_iv_i^*).$$
Now let us assume that the claim is true when the rank of $X$ is $1$. Under this assumption, from $(4)$ we know that for every $1\le i\le r$, there exists $c_i>0$, such that 
$$\mathcal P(v_iv_i^*)\ge c_i v_iv_i^*.$$ 
It follows that for $c:=\min_{1\le i\le r}c_i>0$,
$$\mathcal P(X)\ge c\sum_{i=1}^r v_iv_i^*=cX\Rightarrow {\rm Im}(\mathcal P(X))\supset  {\rm Im}(X).$$
Therefore, it suffices to show that the claim is true when the rank of $X$ is $1$, i.e. $X=vv^*$ for some $v\in\mathbb C^{mn}$. 
Denote $v=(w_1,\cdots,w_n)^t$, where $w_1,\cdots,w_n\in\mathbb C^m$ and $t$ means transpose. By definition, 
$X=(w_iw_j^*)_{n\times n}$, so by $(1)$,
$${\rm tr_1}(X)=\sum_{i=1}^nw_iw_i^*, \quad{\rm tr_2}(X)=(x_{ij})_{n\times n}, \text{ where } x_{ij}= w_j^*w_i.\tag{9}$$
Note that by $(6)$, for $i=1,2$, ${\rm tr}_i(X)\ge 0$, and hence ${\rm Ker}({\rm tr}_i(X))$ and and ${\rm Im}({\rm tr}_i(X))$ are orthogonal to each other. Then it is easy to see that
$${\rm Ker}(\mathcal P(X))= {\rm Ker}({\rm tr_2}(X))\otimes\mathbb C^m + \mathbb C^n\otimes {\rm Ker}({\rm tr_1}(X)).\tag{10}$$
Therefore, due to $(4)$ and $(10)$, it suffices to show that for $u=u_2\otimes u_1\in \mathbb C^n\otimes\mathbb C^m$, if ${\rm tr}_i(X)u_i=0$ for $i=1$ or $2$, then $Xu=0$. 
Denote $u_2=(a_1,\dots,a_n)^t$. Then $u=(a_1u_1,\dots, a_nu_1)^t$, and 
$$Xu=\big((\sum_{j=1}^n a_jw_j^*u_1)w_1,\cdots,(\sum_{j=1}^n a_jw_j^*u_1)w_n\big)^t.\tag{11} $$
From $(9)$ we know that


*

*${\rm Ker}({\rm tr}_1(X))$ is the orthogonal complement of ${\rm span}(w_1,\dots,w_n)$, so if ${\rm tr}_1(X)u_1=0$, then $w_j^*u_1=0$ for $j=1,\dots,n$, and hence $(11)$ is $0$.

*${\rm tr}_2(X)u_2=\big((\sum_{j=1}^n a_jw_j^*)w_1,\cdots,(\sum_{j=1}^n a_jw_j^*)w_n\big)^t$, so if ${\rm tr}_2(X)u_2=0$, $(11)$ is $0$.


The proof of claim is completed. $\quad\square$
