Homework Help: Quantum Information Theory 1

[maomao]

Homework Statement

Given a vector space $X = \mathbb{C}^\Sigma$, with a standard basis $\left\{ |a\rangle, a \in \Sigma \right\}$. Given two operators $V, W \in \mathcal{L}(\mathcal{X})$, the Schur product of $V$ and $W$, denoted by $V \odot W \in \mathcal{L}(\mathcal{X})$, is defined as the operator whose components respect to the standard basis are:
$\langle a| V \odot W |b \rangle = \langle a| V |b \rangle \langle a| W |b\rangle , \forall a, b \in \Sigma$.
Given a positive semidefinite operator $P \in \text{Pos}(\mathcal{X})$, the associated Schur superoperator to $P$, denoted $\Lambda_P \in \text{T}(\mathcal{X})$, is defined as $\Lambda_P (X) = P \odot X, \forall X \in \mathcal{L}(\mathcal{X})$.
Prove that $\Lambda_P (X) \in \text{CP}(\mathcal{X})$.

Relevant Equations

The relevant points I have to prove that a superoperator is completely positive (CP) $\Phi \in \text{T}(\mathcal{X},\mathcal{Y})$, where $\mathcal{X}$ and $\mathcal{Y}$ are complex Euclidean spaces, are the following statements:
1. $\Phi$ is CP, e.g., $\Phi \in \text{CP}(\mathcal{X}, \mathcal{Y})$.

2. $\Phi \otimes \mathbb{I}_{\mathcal{L}(\mathcal{X})}$ is positive.

3. $J(\Phi) \in \text{Pos}(\mathcal{Y} \otimes \mathcal{X})$.

4. $A_1, \cdots , A_N \in \mathcal{L}(\mathcal{X},\mathcal{Y})$ exist so that,
$\Phi (X) = \sum_{k=1}^N A_k X A^{\dagger}_k , \forall X \in \mathcal{L}(\mathcal{X})$

5. Point 4. holds for $N = \text{rank} [J( \Phi)]$.

6. There exists a complex Euclidean space $\mathcal{Z}$ and an operator $A \in \mathcal{L}(\mathcal{X}, \mathcal{Y} \otimes \mathcal{Z})$ such that:
$\Phi (X) = \text{Tr}_{\mathcal{Z}} (AXA^\dagger) , \forall X \in \mathcal{L}(\mathcal{X})$

7. Point 6. holds for $\text{dim} \mathcal{Z} = \text{rank} [J(\Phi)]$.

I considered an operator $X \in \mathcal{L}(\mathcal{X} \otimes \mathcal{K})$, that is positive, $X \geq 0$. And I defined it as it follows:

$X = \sum_{i,j} a_{ij} ∣x_i \rangle \langle x_i ∣ \otimes ∣k_j \rangle \langle k_j∣$​

Where $x_i$ are basis for $\mathcal{X}$ and $k_j$ basis for $\mathcal{Y}$. Then, we remember the Schur product for the operator: $\Lambda_P (X) = P \odot X$.
Another important point is that a superoperator $\Phi$ preserves the trace: $\text{Tr} [\Phi (X)] = \text{Tr} (X), \forall X \in \mathcal{L}(\mathcal{X})$, and since the trace is $\text{Tr}(A) = \sum_i \langle i ∣ A ∣i \rangle$, then,

$\text{Tr} [\Phi (X)] = \text{Tr} [ P \odot X ] = \sum_k \langle k∣ P \odot X ∣k \rangle = \langle k∣ P ∣k \rangle \hspace{1mm} \sum_k \langle k∣ X ∣k \rangle$​

The sum in $k$ represents the Trace of $X$, $\text{Tr}(X)$, and this is why $\langle k∣ P ∣k \rangle = 1$, but I am not sure how could this prove that $\Lambda_P (X)$ is CP.
I understand that, since X is semi-definite positive, $\langle k∣ X ∣k \rangle \geq 0$. Then, if $\langle k∣ P ∣k \rangle$, which I already said that it has to be equal to $1$ to hold the equality, is also non-negative, the product $\langle k∣ P ∣k \rangle \langle k∣ X ∣k \rangle$ will also be non-negative. Is this enough to prove that $\Lambda_P (X) \geq 0$?