# How to prove the inclusion relation $\text{Im} (\rho) \subseteq \text{Im} (\rho[X] \otimes \rho[Y])$ about density operators?

For $$\rho \in \mathrm{D}(\mathcal{X} \otimes \mathcal{Y})$$ denoting an arbitrary state of the pair $$(\mathrm{X}, \mathrm{Y})$$, how to prove the fact
$$\text{Im} (\rho) \subseteq \text{Im} (\rho[X] \otimes \rho[Y])$$ ?

This is an equation (5.109) on page 270 in John Watrous's book named "THE THEORY OF QUANTUM INFORMATION"

• This feels like notational overload. How are $\rho[X]$ and $\rho[Y]$ defined if each $\rho$ is defined over a pair of systems? Commented Apr 28 at 16:32
• $\rho[X]=Tr_{Y}(\rho), \rho[Y]=Tr_{X}(\rho)$, are the partial traces Commented Apr 28 at 21:44
• I suggest starting to think of the special case of $\rho$ being a pure, entangled state, and think about it in the Schmidt basis. What do $\rho[X]$ and $\rho[Y]$ look like? Then think about the spectral decomposition of $\rho$. Commented Apr 29 at 6:43

Let's consider some examples first to see what we're talking about:

• Start thinking about a simple case: what does this mean if $$\sqrt2|\Psi\rangle=|00\rangle+|11\rangle\in\mathbb{C}^2\otimes\mathbb{C}^2$$? Given $$\rho\equiv |\Psi\rangle\!\langle \Psi|$$, the partial states are $$\rho_X = \rho_Y = \frac I2.$$ Therefore $$\operatorname{im}(\rho_X)=\operatorname{im}(\rho_Y) = \mathbb{C}^2$$, and $$\operatorname{im}(\rho_X\otimes\rho_Y)$$ is the just the whole 4-dimensional space. Contrast this with $$\operatorname{im}(\rho)$$, which is the one-dimensional subspace spanned by $$|\Psi\rangle$$ itself. It's clear that the image of $$\rho$$ (as any other subspace) is contained in the image of $$\rho_X\otimes \rho_Y$$ (which is the full space).

• If $$\sqrt3|\Psi\rangle=|00\rangle+\sqrt2|11\rangle$$ is a two-qubit non-fully-entangled state, then $$\rho_X=\rho_Y= \frac13 |0\rangle\!\langle0| + \frac23 |1\rangle\!\langle1|.$$ Again, $$\rho_X,\rho_Y$$ have full-rank images, and thus so is their tensor product, and $$\operatorname{im}(\rho_X\otimes\rho_Y)$$ is again the entire space.

• What if you have instead something like $$\rho=|0\rangle\!\langle0|\otimes I/2$$? In this case $$\rho_X=|0\rangle\!\langle 0|$$, $$\rho_Y=\frac I2$$, and thus $$\rho=\rho_X\otimes\rho_Y$$, and the statement is trivial.

• Let $$2\rho=\mathbb{P}_{00}+\mathbb{P}_{11}$$, $$\mathbb{P}_{ij}\equiv |ij\rangle\!\langle ij|$$. Now $$\rho_X=\rho_Y = I/2$$, and thus $$\rho_X\otimes\rho_Y=I/4$$ is the identity matrix, whose image is again the full space $$\mathbb{C}^4$$. Contrast this with $$\operatorname{im}(\rho) = \operatorname{span}(\{|00\rangle,|11\rangle\}),$$ and we again see that the statement is trivially true.

As for the general statement:

• A bit more generally, given an arbitrary pure bipartite state $$|\Psi\rangle=\sum_i \sqrt{p_i} |u_i,v_i\rangle\in\mathbb{C}^n\otimes\mathbb{C}^m$$, for some pair of orthonormal bases $$\mathcal U\equiv \{|u_i\rangle\}_i$$ and $$\mathcal V\equiv \{|v_i\rangle\}_i$$, then the images of the reduced states are spanned by $$\mathcal U$$ and $$\mathcal V$$, and thus the image of $$\rho_X\otimes\rho_Y$$ by vectors of the form $$|u_i,v_j\rangle$$ for all $$i,j$$.

In particular, that means that if $$|\Psi\rangle$$ has full Schmidt rank, then $$\operatorname{im}(\rho_X\otimes\rho_Y)=\mathbb{C}^n\otimes\mathbb{C}^m$$ is the full space, making the statement automatically true. And even if the image of $$\rho_X\otimes \rho_Y$$ doesn't span the full space, $$|\Psi\rangle$$ is always in it, because $$|u_i,v_i\rangle\in\operatorname{im}(\rho_X\otimes\rho_Y)$$ for all $$i$$. More formally, we showed that $$|u_i\rangle\in\operatorname{im}(\rho_X)$$, $$|v_i\rangle\in\operatorname{im}(\rho_Y)$$, thus $$|u_i,v_i\rangle\in\operatorname{im}(\rho_X\otimes\rho_Y)$$, and thus $$|\Psi\rangle \in \operatorname{im}(\operatorname{tr}_Y\mathbb{P}_{\Psi} \otimes\operatorname{tr}_X \mathbb{P}_{\Psi}).$$

• Finally, let $$\rho$$ be a generic state. Its image is always the span of its eigenvectors, so you can reduce the question to: are all eigenvectors of $$\rho$$ in the image of $$\rho_X\otimes\rho_Y$$? To answer in the positive, we observe that if $$\rho=\sum_i p_i \mathbb{P}_{\Psi_i}$$ for some set of (generally entangled) orthonormal states $$|\Psi_i\rangle$$, then $$\rho_X = p_1 \operatorname{tr}_Y[\mathbb{P}_{\Psi_1}] + (\text{positive stuff}).$$ In words, we're saying that doing the partial trace of (projections onto) eigenstates only ever gives positive semidefinite operators. But also, for any pair of positive semidefinite Hermitians $$P,Q\ge0$$, we know that $$P+Q\ge Q$$, and that if $$Q\le P$$ then $$\operatorname{im}(Q)\subseteq \operatorname{im}(P)$$. It follows that $$|\Psi_1\rangle \in \operatorname{im}(\operatorname{tr}_Y\mathbb{P}_{\Psi_1} \otimes\operatorname{tr}_X \mathbb{P}_{\Psi_1}) \subseteq \operatorname{im}(\rho_X\otimes\rho_Y),$$ where I used $$\operatorname{tr}_Y\mathbb{P}_{\Psi_1} \le \rho_X, \qquad \operatorname{tr}_X\mathbb{P}_{\Psi_1} \le \rho_Y.$$ The same identical reasoning can be applied to show that all eigenvectors $$|\Psi_i\rangle$$ are in the image of $$\rho_X\otimes\rho_Y$$, thus the conclusion.