Kind of a nooby question. But how would you actually go about proving that the Werner Holevo channel is a quantum channel?
$$ C(X) = \frac{1}{n \pm 1} \left(\text{Tr}(X)\mathbb{I}_n \pm X^{\top} \right)$$ Where $n$ in the dimension of the map.
I know that I have to find that it is linear, trace preserving and completely positive. I tried to rewrite it so that I sum over the element indices but it didn't help me much.
The mapping is clearly linear and trace preserving, so it remains to show that it is completely positive.
The best way to see this is to use the Choi representation of the mapping.
Given a mapping $\Phi:M_n\to M_m$ (where $M_n$ and $M_m$ are the spaces of $n\times n$ and $m\times m$ matrices respectively), its Choi representation is the operator $J(\Phi)\in M_m\otimes M_n$ defined by $$ J(\Phi) = \sum_{i,j=1}^n \Phi(E_{i,j})\otimes E_{i,j} $$ where for each pair of indices $i,j\in\{1,\dots n\}$, one defines $E_{i,j}$ as the $n\times n$ matrix whose $(i,j)$-entry is equal to 1 and the rest of the entries are zero. It is known that $\Phi$ is completely positive if and only if $J(\Phi)$ is positive semidefinite as $(nm)\times(nm)$ matrix.
The Choi representation of the identity mapping $\mathrm{id}_{M_n}$ is simply $$ J(\mathrm{id}_{M_n}) = \sum_{i,j=1}^n E_{i,j}\otimes E_{i,j} = uu^* $$ where $u$ is the (unnormalized) maximally entangled state vector $$ u = \sum_{i=1}^n e_{i}\otimes e_{i}. $$ The Choi representation of the transpose mapping $T:M_n\to M_n$ is $$ J(T) = \sum_{i,j=1}^n E_{i,j}^\top\otimes E_{i,j} = \sum_{i,j=1}^n E_{j,i}\otimes E_{i,j} = (T\otimes \mathrm{id})(uu^*) $$ where $(T\otimes \mathrm{id})$ is the partial transpose mapping.
Now, in your case, we have \begin{aligned} J(C) &= \frac{1}{n\pm 1} \left(\mathbb{I}\otimes \mathbb{I} \pm (T\otimes \mathrm{id})(uu^*)\right)\\ %& =\frac{1}{n\pm 1}(T\otimes\mathrm{id})\left(\mathbb{I}\otimes \mathbb{I} \pm uu^*\right). \end{aligned} Can you show that this is positive semidefinite?