Eigenvalues of a reduced density matrix

[barnflakes]

My lecturer keeps telling me that if a density matrix describes a pure state then it must contain only one non-zero eigenvalue which is equal to one. However I can't see how this is true, particularly as I have seen a matrix \rho_A = \begin{pmatrix} 1/2 & - 1/2 \\ -1/2 & 1/2 \\ \end{pmatrix} for which this is not true. He then clarified that if it was in "the diagonal basis" this was true. Can someone clarify this for me or show me a proof please?

[Tomsk]

I guess he means that you can write a density matrix as
\rho = | \psi_i \rangle \langle \psi_i |
Which has eigenvalues \delta_{ij} since
\rho |\psi_j \rangle = \delta_{ij} |\psi_i \rangle.
The example you gave can be written as \rho = | - \rangle \langle - |, where | - \rangle = \frac{1}{\sqrt{2}} \begin{pmatrix}1 \\ -1\end{pmatrix} which has | - \rangle as an eigenvector, with eigenvalue 1.

[barnflakes]

Is that true? I thought $\psi_i \rangle$ and $\psi_j \rangle$ weren't necessarily orthogonal. He also definitely said that when it's pure it only has one non-zero eigenvalue. Maybe he means only in the basis of its eigenvectors, I don't know? That's what I'm trying to find out.

[Tomsk]

Yeah I think it is the basis of its eigenvectors. I think you can always take a pure state as one possible basis vector. E.g. if you have |\psi_a \rangle = \alpha |0 \rangle + \beta |1 \rangle then the other one is |\psi_b \rangle = \alpha |0 \rangle - \beta |1 \rangle. I can't remember if/how this generalizes to higher dimensions though. :blushing: Maybe Gram-Schmidt?