QuantumInfo.Channels.Unbundled

Let $R$ be a semiring and $\{\Lambda_i\}_{i \in I}$ be a family of linear matrix maps $\Lambda_i : \text{Mat}_{dI_i}(R) \to \text{Mat}_{dO_i}(R)$. If each map $\Lambda_i$ is trace-preserving, then their tensor product $\bigotimes_{i \in I} \Lambda_i$ is also a trace-preserving matrix map.

Let $A$ and $B$ be finite types. Let $M: \text{Mat}_A(\mathbb{C}) \to \text{Mat}_B(\mathbb{C})$ be a completely positive linear map between complex matrix spaces. If $M$ is subunital, meaning $M(I) \le I$ where $I$ is the identity matrix, then for any matrix $X \in \text{Mat}_A(\mathbb{C})$, the Kadison-Schwarz inequality holds: $$(M(X))^\dagger M(X) \le M(X^\dagger X)$$ where $X^\dagger$ denotes the conjugate transpose of $X$ and the inequality $\le$ refers to the Loewner order (positive semidefiniteness).

Let $M: \text{Mat}_A(\mathbb{C}) \to \text{Mat}_B(\mathbb{C})$ be a positive linear map between spaces of complex square matrices. If $M$ is subunital, satisfying $M(I) \le I$ where $I$ is the identity matrix, then for any positive semi-definite matrix $X \ge 0$, the operator norm of its image is bounded by the operator norm of the input: $\|M(X)\| \le \|X\|$.

Let $A$ and $B$ be finite types. Let $M: \text{Mat}_A(\mathbb{C}) \to \text{Mat}_B(\mathbb{C})$ be a linear map between complex square matrix spaces. If $M$ is completely positive and subunital (that is, $M(I_A) \le I_B$, where $I$ denotes the identity matrix and $\le$ denotes the Loewner order), then for any matrix $X \in \text{Mat}_A(\mathbb{C})$, the operator norm of $M(X)$ is less than or equal to the operator norm of $X$: $$\|M(X)\| \le \|X\|$$