# Is there an easy way to calculate the eigenvalues of the partial transpose of a given matrix? [duplicate]

Consider the state $$|\psi\rangle=(\cos\theta_A|0\rangle+\sin\theta_A|1\rangle)\otimes(\cos\theta_B|0\rangle+e^{i\phi_B}\sin\theta_B|1\rangle).$$

To calculate the $$\rho^{T_B}$$ I first calculate the $$\rho$$ and the change it into a matrix form but It was very complicated to compute the eigenvalues for that matrix.

Is there any easy way to calculate the eigenvalues of $$\rho^{T_B}$$?

In this specific case, absolutely! Note that $$|\psi\rangle=|\phi_A\rangle|\phi_B\rangle,$$ such that $$\rho=|\phi_A\rangle\langle\phi_A|\otimes |\phi_B\rangle\langle\phi_B|.$$ Now, it is the case that $$|\phi_B\rangle\langle\phi_B|^T=|\phi_C\rangle\langle\phi_C|$$ for some state $$|\phi_C\rangle=\cos\theta_B|0\rangle+e^{-i\phi_B}\sin\theta_B|1\rangle$$.
So, you should now be able to simply observe that $$\rho^{T_B}=|\phi_A\rangle\langle\phi_A|\otimes |\phi_C\rangle\langle\phi_C|$$ is a rank 1 projector ($${\rho^{T_B}}^2=\rho^{T_B}$$ and $$\text{Tr}(\rho^{T_B})=1$$), or that the only non-zero eigenvector is just $$|\phi_A\rangle|\phi_C\rangle$$, and that it has eigenvalue 1.