Understanding Quantum Purification

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I am not very good at this, but I really want to understand it.

Definition:
We say lψ>AB is a purification of ρA if TrB[lψ><ψlAB ] = ρA.

Note that ρA is a density matrix.

My book proceeds to give an example of a purifying system as:

√ρA ⊗ 1*lΦ>, where 1 is identity and lΦ> = ∑ilii> = ∑ili>⊗li>

How can I see that this is true? I am not sure I even know how to perform the tensor product on the LHS. Is this correct?
√ρA ⊗ 1*lΦ> = √ρA ⊗ (∑ili>⊗li>) = (∑i√ρAli>⊗√ρAli>)
 
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That's definitely not correct. The matrix sizes of the left-hand and right-hand sizes can't possibly match up because you took ##\rho_A## and suddenly put it on both sides of the tensor product. I think you're confused about what your book is doing. The starting expression ##\sqrt{\rho_A} \otimes \left| \Phi \right\rangle## already makes no sense in context. Like... it's tensor-producting a matrix against a column-vector.

To do a purification on paper you generally:

- Perform an eigendecomposition of the density matrix ##\rho##, giving you ##\rho = \sum p_i \left| v_i \right\rangle \left\langle v_i \right|## where each ##p_i## is a probability and each ##\left| v_i \right\rangle## is some pure state (the ##p_i##'s are the eigenvalues and the ##\left| v_i \right\rangle##'s are the eigenvectors).
- Introduce a secondary pure state which has a state for each eigenvector and amplitudes that give probabilities that will match the eigenvalues. Perhaps ##\sum \sqrt{p_i} \left| i \right\rangle##.
- Match the secondary pure state's components against the density matrix' eigenvectors, inside the sum. ##\psi_{pure} = \sum \sqrt{p_i} \left| i \right\rangle \left| v_i \right\rangle##.
- The new overall density matrix is ##\rho_{pure} = \sum \sum \sqrt {p_i} \sqrt{p_j} \left| i \right\rangle \left| v_i \right\rangle \left\langle v_j \right| \left\langle j \right|##, and tracing over the introduced secondary state's space will result in the original density matrix.

This is exactly what the wikipedia page on quantum purification does.
 
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