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Consider the CPTP map $\Phi: M_n(\mathbb{C})\rightarrow M_n(\mathbb{C})$ that is not injective, which means the null space of $\Phi$ is non-trivial. My main question is, what can we say about the matrices in this null space?

By this question, we can take a basis of $M_n(\mathbb{C})$, denoted by $\{D_1,D_2,\dots,D_{n^2}\}$ where $D_i$'s are all $n\times n$ density matrices. Since $\Phi$ is not injective, we can get that $\operatorname{span}\{\Phi(D_1),\Phi(D_2),\dots,\Phi(D_{n^2})\}$ is a proper subspace of $M_n(\mathbb{C})$. In other words, $\Phi(D_1),\Phi(D_2),\dots,\Phi(D_{n^2})$ are not linearly independent. So we have that $$ \Phi(\alpha_1 D_1 + \alpha_2 D_2 + \cdots \alpha_{n^2} D_{n^2})=\alpha_1 \Phi(D_1) + \alpha_2 \Phi(D_2) + \cdots + \alpha_{n^2} \Phi(D_{n^2})=0 $$ for some $\alpha_i$'s that are not all zero. On the other hand, since $D_1,D_2,\dots,D_{n^2}$ are linearly independent, this implies that $\alpha_1 D_1 + \alpha_2 D_2 + \cdots \alpha_{n^2} D_{n^2}$ is a non-trivial element in the null space of $\Phi$.

Now let's look at this non-trivial element. It is clearly Hermitian, and its trace is zero. Of course, we can also use it to span a one-dimensional subspace of the null space. Is that all we can say about the matrices in the null space of $\Phi$?

Edit: This proof may be not sufficient. But the above conclusion should be still correct, by the proof that I provided in the comments.

Thinkpad
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  • Other than the null space having vanishing trace, I don't think you can say much else about the null space in general. – Luftbahnfahrer Jun 18 '23 at 14:21
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    Note that $M_n$ is a complex vector space, so the coefficients $\alpha_i$ can in general be complex, so the null space isn't Hermitian. Unless you are interested in the real vector space of nxn Hermitian matrices. Which also has dimension $n^2$. – Luftbahnfahrer Jun 18 '23 at 14:23
  • @Luftbahnfahrer Thanks, I think you are right. But I still want to claim that there must be Hermitian matrices in the null space, by another proof. Consider a non-trivial element $A$ such that $\Phi(A)=0$. We can decompose $A$ into $\frac{A+A^}{2}$ and $\frac{A-A^}{2}$, which are Hermitian and skew-Hermitian, respectively. Note that the quantum channel $\Phi$ preserves Hermitianess and skew-Hermitianess. So we get that $\Phi(\frac{A+A^}{2})$ and $\Phi(\frac{A-A^}{2})$ are Hermitian and skew-Hermitian, respectively. – Thinkpad Jun 19 '23 at 03:07
  • @Luftbahnfahrer Since $\Phi(\frac{A+A^}{2})+\Phi(\frac{A-A^}{2})=\Phi(A)=0$, it implies that $\left(\Phi\left(\frac{A+A^}{2}\right)+\Phi\left(\frac{A-A^}{2}\right)\right)^=\Phi\left(\frac{A+A^}{2}\right)-\Phi\left(\frac{A-A^}{2}\right)=0$. These two equalities give that one of $\frac{A+A^}{2}$ and $\frac{A-A^}{2}$ is in the null space and is non-trivial as well. Note that $\frac{A+A^}{2}$ and $i\frac{A-A^*}{2}$ are both Hermitian, which completes the proof. – Thinkpad Jun 19 '23 at 03:07
  • @Luftbahnfahrer Do you think this is all we can do? I wish there could be more information regarding the eigenvalues of this Hermitian matrix. – Thinkpad Jun 19 '23 at 03:07
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    Since channels are Hermitian-preserving maps, then yes, as you observed, if the null space is nontrivial then it must contain a nontrivial Hermitian element. I don't think you can say much about the eigenvalues of null-space elements, though, without more specifics about the mapping. – Luftbahnfahrer Jun 21 '23 at 14:30

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