# Does a quantum channel map the maximally mixed input state into an output state with maximal rank?

Consider a quantum channel $$\Phi : M_n \rightarrow M_m$$ and let $$\frac{\mathbb{I}_n}{n}$$ be the maximally mixed input state. For all input states $$\rho\in M_n$$, is it true that

$$\quad \text{rank} \, \Phi (\rho) \leq \text{rank}\, \Phi \left(\mathbb{\frac{I_n}{n}}\right)?$$

Let's forget about normalization of the two sides of the inequality as if $$\operatorname{rank}(A) \leq \mathrm{rank}(B)$$ then $$\mathrm{rank}(c A) \leq \mathrm{rank}(d B)$$ for any $$c,d \neq 0$$.

Now let's write $$I_n = \rho + (I-\rho)$$. As $$\rho \leq I_n$$ because $$\rho$$ is a quantum state, we have $$(I-\rho) \geq 0$$. Then \begin{aligned} \Phi(I_n) &= \Phi(\rho + (I_n-\rho)) \\ &= \Phi(\rho) + \Phi(I_n-\rho) \end{aligned} Thus $$\mathrm{rank}(\Phi(I_n)) = \mathrm{rank}(\Phi(\rho) + \Phi(I_n-\rho))$$. But as $$\Phi(\rho)$$ and $$\Phi(I_n - \rho)$$ are both positive-semidefinite we have $$\mathrm{rank}(\Phi(\rho) + \Phi(I_n-\rho)) \geq \mathrm{rank}(\Phi(\rho))$$ (see this answer for a proof of that).

Putting it all together we get $$\mathrm{rank}(\Phi(I_n)) \geq \mathrm{rank}(\Phi(\rho) )$$ which implies the idenity in the question.

Regarding the title of the question You should not interpret this result as saying that the maximally mixed state is always mapped to a state of maximal rank in the output system. For example, you can take take a channel $$\Phi(\rho) = \mathrm{Tr}(\rho) |0\rangle \langle 0 |$$ whose output will always have rank $$1$$ regardless of the dimension of the output sysytem.

Yes, since for an arbitrary state $$\rho \in M_n$$, there exists a $$\delta > 0$$ such that $$\delta \rho \leq \mathbb{I_n}/n$$. The positivity of the channel then implies that $$\Phi (\delta \rho) = \delta \Phi(\rho) \leq \Phi (\mathbb{I_n}/n)$$, which clearly implies that $$\text{rank}\, \Phi (\rho) \leq \text{rank}\, \Phi (\mathbb{I_n}/n).$$

Consider the Kraus representation of the channel $$\Phi:\mathrm{Lin}(\mathcal X)\to\mathrm{Lin}(\mathcal Y)$$: $$\Phi(X)=\sum_{a=1}^d A_a X A_a^\dagger,\tag1$$ for all $$X\in\mathrm{Lin}(\mathcal X)$$, for some positive integer $$d$$ and set of $$d$$ linear operators $$A_a\in\mathrm{Lin}(\mathcal X,\mathcal Y)$$ such that $$\sum_a A_a^\dagger A_a=I_{\mathcal X}$$.

Note that $$\Phi(I_{\mathcal X}) = \sum_{a=1}^d A_a A_a^\dagger$$. Therefore $$\operatorname{supp}(\Phi(I_{\mathcal X}))=\bigcup_a \operatorname{supp}(A_a),$$ and $$\operatorname{supp}(\Phi(X))\subseteq \operatorname{supp}(\Phi(I_{\mathcal X}))$$.