# If $\rho_{AB}$ is a separable then the partial transpose w.r.t to A is PSD

Def: The partial transpose of a linear operator $$\rho_{AB}$$ over a Hilbert space $$H_A \otimes H_B$$ w.r.t A is defined for a linear operator $$\rho_{AB}=\rho_A \otimes\rho_B$$ as $$\rho^{T_A}_{AB}=\rho_A^T \otimes\rho_B$$ The definition can be extended to a general linear operator

I want to prove the Partial transpose test: If $$\rho_{AB}$$ is a separable (unentangled) then the partial transpose w.r.t to A is PSD (positive semidefinite)

My try:

My definition of PSD is that a hermitian operator is PSD if it has non negative eigenvalues

Assume that $$\rho_{AB}=\rho_A \otimes\rho_B$$. Then since $$\rho_A^T$$ and $$\rho_A$$ have the same eigenvalues, then they both have non-negative eigenvalues. I think I can say they are both PSD but they still have to be hermitian for that and it looks like the transpose is not hermitian: $$(\rho_A^T)^\dagger=\overline{(\rho_A^T)^T}=\overline{\rho_A}$$.

Furthermore how do I conclude that $$\rho^{T_A}_{AB}=\rho_A^T \otimes\rho_B$$ is hermitian and with nonnegative eigenvalues? I don't know what the eigenvalues of a tensor product are in terms of the eigenvalues of the initial spaces.

Finally I have to extend this to the general case. If $$\rho_{AB}$$ is a generic separable state then it a convex linear combination of product states: $$\rho^{T_A}_{AB}=\sum_{i=1}^d c_i \rho_{A_i}^T \otimes\rho_{B_i}$$ And then the eigenvalues are the sum of the eigenvalues of the addends, so they are non negative .To prove it is PSD again I need to prove that it is hermitian but I guess that if the addends $$\rho_A^T \otimes\rho_B$$ are prove to be hermitian, then a convex linear combination of hermitian matrices is hermitian

How do I complete or fix the proof?

The transpose of a Hermitian matrix is again Hermitian, and your attempt got quite close. All you have to do is, essentially, the same calculation again: $$(\rho^T)^\dagger=\overline{(\rho^T)^T}=\overline{\rho}=\overline{\rho^\dagger}=\overline{\overline{\rho^T}}=\rho^T$$ (In the third step we used that $$\rho$$ is Hermitian)