I am reading Introduction to Quantum Computing by Kaye, Laflamme, and Mosca. Here is a question I am struggling with:
Exercise 3.5.6: Prove that the transpose map, which maps $\rho \mapsto \rho^{T}$ is positive1, but not completely positive2.
Now, positivity is defined in terms of the inner product, i.e. $\rho$ is positive iff $\forall v\in \mathcal{H}$, $\left<v, \rho v\right> \ge 0$, but "transpose" is defined in terms of an operation on matrices. So I was able to get this under the assumption that $\rho^{T} = \rho^{*}$, but not under the weaker assumption the $\rho$ is simply positive. Is this true even if I don't assume $\rho^{T} = \rho^{*}$?
As for showing that the transpose map is not completely positive, I frankly do not know what I'm doing and am asking for any and all help you can give me. My attempt is given below, though it is not worth reading:
Let $\rho\otimes \gamma$ be a positive map, so that $\forall u\otimes v \in \mathcal{H}_A \otimes \mathcal{H}_{B}$ $$ \left<u\otimes v, \rho u \otimes \gamma v\right> := \left<u, \rho u\right>\left< v, \gamma v \right> \ge 0. $$ Then the transpose map tensored with the identity takes $\rho \otimes \gamma$ to $\rho^{T}\otimes \gamma$, and we have (again assuming $\rho^{T}=\rho^{*}$) $$ \left<u\otimes v, \rho u \otimes \gamma v\right> := \left<u, \rho^{T} u\right>\left< v, \gamma v \right> = \overline{\left< u, \rho u\right>}\left<v, \gamma v \right> =\frac{\overline{\left< u, \rho u\right>}}{\left< u, \rho u\right>}\left<u, \rho u\right>\left< v, \gamma v \right>. $$ So $\frac{\overline{\left< u, \rho u\right>}}{\left< u, \rho u\right>}$ must be $\ge 0$. . .(at this point I am stumped.)
1 "Positive" in this context means "maps positive operators to positive operators.
2 They define completely positive as follows: A map is completely positive iff it is positive, and in addition, when tensored with the identity operation, they still map positive operators to positive operators.