Why can input states $\rho_m$ be chosen to be rank-one operators in Zero-error communication? I have one problem in Quantum Information Theory, specially in Zero-error communication.
I am reading a paper :
Duan, Runyao; Severini, Simone; Winter, Andreas, Zero-error communication via quantum channels, noncommutative graphs, and a quantum Lovász number, IEEE Trans. Inf. Theory $59$, No. $2$, $1164-1174$ $(2013)$. ZBL$1364.81059$.
And in page $5$, there is a paragraph : 

Let $N : L(A) \to L(B)$ be a quantum channel, i.e. a linear completely positive, trace-preserving map with Kraus operators $E_j : A \to B$, so that $N(\rho)$ = $\sum_j E_jρE^\dagger_j$. Then to send messages $m$ one has to associate them with states $\rho$ such that different states $\rho, \sigma$ lead to orthogonal channel output states: $N(\rho) \perp N(\sigma)$, because it is precisely the orthogonal states that can be distinguished with certainty. Clearly, these states may, w.l.o.g., be taken as pure (rank 1 operator $\textbf{vv}^*$), as the orthogonality is preserved when going to any states in the support (i.e., the range) of $\rho, \sigma$, etc.

I totally don't understand why the last sentence is clear.
I knew that $N(\rho) \perp N(\sigma)\Leftrightarrow \operatorname{Tr}[N(\rho)N(\sigma)] = 0$,but didn't obtain more informations from $\operatorname{Tr}[N(\rho)N(\sigma)]=0$
Please explain why $\rho, \sigma$ can be replaced by pure states.
 A: We can see why by understanding the following claim.
For a density operator $\rho\in\mathcal{S}(A)$, one may denote the image (or range) of $\rho$ as $\mathrm{im}(\rho)$. Note that, for any vector $u\in\mathrm{im}(\rho)$, there exists a positive real number $x$ such that $uu^*\leq x\rho$. Moreover, for any positive linear mapping $\mathcal{N}$, it holds that $\mathcal{N}(uu^*)\leq x \mathcal{N}(\rho)$.
Claim. Let $\mathcal{N}:\mathcal{L}(A)\rightarrow \mathcal{L}(B)$ be a channel and let $\rho,\sigma\in\mathcal{S}(A)$ be density operators satisfying $\mathrm{Tr}(\mathcal{N}(\rho)\mathcal{N}(\sigma))=0$. For any vectors $u\in\mathrm{im}(\rho)$ and $v\in\mathrm{im}(\sigma)$, it holds that $\mathrm{Tr}(\mathcal{N}(uu^*)\mathcal{N}(vv^*))=0$.
Proof. From the above observations, there exist positive real numbers $x$ and $y$ satisfying $uu^*\leq x\rho$ and $vv^*\leq y\sigma$. It follows that $\mathcal{N}(uu^*)\leq x \mathcal{N}(\rho)$ and $\mathcal{N}(vv^*)\leq y \mathcal{N}(\sigma)$. Moreover,
\begin{align*}
0\leq\mathrm{Tr}(\mathcal{N}(uu^*)\mathcal{N}(vv^*)) \leq xy\,\mathrm{Tr}(\mathcal{N}(\rho)\mathcal{N}(\sigma))\ = 0,
\end{align*}
which yields the desired result.

This idea can be extended to any state in the image of $\rho$ and $\sigma$. Namely, if $\rho_0$ and $\sigma_0$ are any other density operators satisfying $\mathrm{im}(\rho_0)\subseteq\mathrm{im}(\rho)$ and $\mathrm{im}(\sigma_0)\subseteq\mathrm{im}(\sigma)$, there must exist positive constants $x$ and $y$ such that $\rho_0\leq x\rho$ and $\sigma_0\leq y\sigma$ and we find that
\begin{align*}
0\leq\mathrm{Tr}(\mathcal{N}(\rho_0)\mathcal{N}(\sigma_0)) \leq xy\,\mathrm{Tr}(\mathcal{N}(\rho)\mathcal{N}(\sigma))\ = 0.
\end{align*}
