Proof of Nielsen's theorem (Theorem 12.15) given in Nielsen-Chuang (assumption of invertibility)

Question

Theorem 12.15 of Nielsen and Chuang's 10th anniversary edition is Nielsen's Theorem (1999). In particular, it says,

Theorem 12.15: A bipartite pure state $\mid \psi \rangle$ may be transformed to another pure state $\mid \phi \rangle$ by LOCC if and only if $\lambda_\psi \prec \lambda_\phi$.

Here, $\mid \psi \rangle$ is an entangled pure state in the composite system $AB$ and $\lambda_\psi$ denotes the vector of eigenvalues of $\rho_{\psi} = Tr_B(\mid \psi \rangle \langle \psi \mid)$ (which is the density matrix for the state of A's system).

In the proof of the converse, the authors assume that $\rho_{\psi}$ is invertible in order to define measurement operations $M_j = \sqrt{p_j \rho_\phi}U_j^{\dagger}\rho_{\psi}^{-1/2}$ for Alice's system. They say "this assumption is easily removed; see Exercise 12.20" but that exercise doesn't give any indication on how to remove this assumption.

Exercise 12.20: Show that the assumption that $\rho_{\psi}$ is invertible may be removed from the proof of the converse part of Theorem 12.15.

It is not clear to me how to define appropriate measurement operators if we do not have invertibility of $\rho_{\psi}$. My question is, how can this assumption be removed?

Edit: I have added a picture (below) of the full proof given in Nielsen-Chuang, in case this helps clarify how to proceed when we remove the assumption of invertibility.

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Cross-posted on physics.SE

Answer 1

1. Given an arbitrary matrix $A$, with singular value decomposition $A=\sum_k s_k |u_k\rangle\!\langle v_k|$, $s_k>0$, you can always define its pseudo-inverse as $A^+=\sum_k \frac{1}{s_k} |v_k\rangle\!\langle u_k|$. This is kind of an inverse of $A$. If $A$ is invertible, then $A^+=A^{-1}$. If $A$ is injective, then $A^+ A=I$, i.e. $A^+$ is a left inverse. If $A$ is surjective, then $AA^+=I$, i.e. $A^+$ is a right inverse (notice that the identities in these two relations are generally different, if $A$ is not squared). If $A$ is injective you can write it explicitly as $A^+=(A^T A)^{-1}A^T$, while if it's surjective as $A^+=A^T (AA^T)^{-1}$. More generally, $A$ needs not be injective nor surjective, and in these cases $A{A}^+$ is always the projection onto the range of $A$, while $A^+A$ the projection onto its support.

2. We can simply use the pseudoinverse instead of the inverse here. Define $M_j=\sqrt{p_j} \sqrt{\rho_\varphi} U_j^\dagger \sqrt{\rho_\psi^+}$. Notice that $\rho_\psi$ is squared, Hermitian, and not invertible, thus $\rho_\psi\rho_\psi^+=\rho_\psi^+\rho_\psi$ are both the projection onto the support of $\rho_\psi$, and furthermore $\rho_\psi^+$ is Hermitian (you see it immediately writing the pseudoinverse in terms of the singular value decomposition, which amounts to the eigenvalue decomposition for Hermitian operators). You have $$\sum_j M_j^\dagger M_j = \sum_j p_j \sqrt{\rho_\psi^+} U_j \rho_\varphi U_j^\dagger \sqrt{\rho_\psi^+} = \sqrt{\rho_\psi^+} \rho_\psi \sqrt{\rho_\psi^+}.$$

3. For any $A$, you can readily see writing explicitly the terms via the singular value decomposition that $\sqrt{A^+} A \sqrt{A^+}$ equals the projection onto the range of $A$. Thus $\sum_j M_j^\dagger M_j$ is the projection onto the support (equivalently, the range) of $\rho_\psi$.

4. We thus found that $\sum_j M_j^\dagger M_j$ is not quite the identity. However, it's close enough. You can easily make it an identity by defining the action of $M_j$ onto the subspace orthogonal to the support of $\rho_\psi$ in an arbitrary way that ensures the overall normalization is satisfied. For example, you could simply define $$M_j = \sqrt{p_j}\sqrt{\rho_\varphi}U_j^\dagger \sqrt{\rho_\psi^+} + \frac{\Pi_{\operatorname{supp}(\rho_\psi)_\perp} }{ \sqrt m },$$ where $m$ is the number of measurement outcomes, and $\Pi_{\operatorname{supp}(\rho_\psi)_\perp}$ is the projection onto the subspace orthogonal to the support of $\rho_\psi$.