Does the tensor product respect semidefinite ordering in this way? I'll use $\succeq$ to denote the positive semidefinite ordering: for square matrices $X,Y$, one has $X \succeq Y$ iff $X - Y$ is positive semidefinite. It's a well known fact that if $X, Y \succeq 0$ then $X \otimes Y \succeq 0$. However, if one has two pairs of matrices with $X \succeq Y$ and $X' \succeq Y'$, it isn't necessarily the case that $X \otimes X' \succeq Y \otimes Y'$ (for instance, if $X = X' = I$ and $Y = Y' = -2I$).
My question is: if we add that $Y, Y' \succeq 0$, so our assumptions are
  $$ X \succeq Y \succeq 0,\ \ \ X' \succeq Y' \succeq 0 $$
Is it necessarily true that $X \otimes X' \succeq Y \otimes Y'$?
I personally feel that this should be true (and I could swear I've seen this result before but can't find it anywhere), but I'm struggling to prove it. Does anyone have a reference for this fact, and/or know how to prove (or disprove) it?
 A: Yes, it's true. Let $\vec v = (v_1^T,\ldots,v_n^T)^T$ with $v_1,\ldots,v_n\in\mathbb R^n$. Then
$$
\langle (X\otimes X')\vec v,\vec v\rangle = \sum_{i=1}^n\sum_{j=1}^nx_{ij}\langle X'v_j,v_i\rangle.
\quad\text{and}\quad
\langle (Y\otimes Y')\vec v,\vec v\rangle = \sum_{i=1}^n\sum_{j=1}^ny_{ij}\langle Y'v_j,v_i\rangle.
$$
Now, define the $n\times n$ matrices
$$
A = (\langle X'v_j,v_i\rangle)_{i,j=1}^n
\quad\text{and}\quad
B = (\langle Y'v_j,v_i\rangle)_{i,j=1}^n.
$$
Then $\langle (X\otimes X')\vec v,\vec v\rangle = \sum_{i=1}^n\sum_{j=1}^nx_{ij}a_{ji} = \sum_{i=1}^n(XA)_{ii} = \operatorname{tr}(XA)$. Similarly, $\langle (Y\otimes Y')\vec v,\vec v\rangle = \operatorname{tr}(YB)$. We have
$$
w^TAw = \sum_{i,j}a_{ij}w_iw_j = \sum_{i,j}\langle X'(w_jv_j),(w_iv_i)\rangle = \langle X'z,z\rangle,\quad w^TBw = \langle Y'z,z\rangle,
$$
where $z = \sum_iw_iv_i$, and therefore, $A\ge B$. Hence,
\begin{align}
\langle (X\otimes X')\vec v,\vec v\rangle
&= \operatorname{tr}(XA) = \operatorname{tr}(X^{1/2}AX^{1/2}) = \sum_i\langle AX^{1/2}e_i,X^{1/2}e_i\rangle\\
&\ge \sum_i\langle BX^{1/2}e_i,X^{1/2}e_i\rangle = \operatorname{tr}(XB) = \operatorname{tr}(B^{1/2}XB^{1/2})\\
&\ge \operatorname{tr}(YB) = \langle (Y\otimes Y')\vec v,\vec v\rangle.
\end{align}
A: \begin{aligned}
&X\otimes X'-Y\otimes Y'\\
&=\left[(X-Y)+Y\right]\otimes\left[(X'-Y')+Y'\right]-Y\otimes Y'\\
&=\left[(X-Y)\otimes(X'-Y')+Y\otimes(X'-Y')+(X-Y)\otimes Y'+Y\otimes Y'\right]-Y\otimes Y'\\
&=(X-Y)\otimes(X'-Y')+Y\otimes(X'-Y')+(X-Y)\otimes Y'.
\end{aligned}
Now the result follows because $X-Y,\,X'-Y',\,Y$ and $Y'$ are all positive semidefinite (and this is where $Y,Y'\succeq0$ is essential).
