Let $ρ_A = \sum_{i=1}^r p_i|e_i⟩⟨e_i|$, where $p_i$ are the nonzero eigenvalues of $ρ_A$ and $|e_i⟩$ corresponding orthonormal eigenvectors. If some eigenvalue appears more than once then this decomposition is not unique.
Show that, nevertheless, any purification $|\psi_{AB}⟩$ of $ρ_A$ has a Schmidt decomposition of
the form $|\psi_{AB}⟩ = \sum_{i=1}^r s_i|e_i⟩ ⊗ |f_i⟩$,...(*)
with the same $|e_i⟩$ as above. Hint: Start with an arbitrary Schmidt decomposition and rewrite it in the desired form.
My try:
Following the hint,let $|\Psi_{AB}\rangle\in H_A\otimes H_B$ be a purification of $\rho_A$ with $H_B$ of dimension
dim$H_B \ge $rank $\rho_A$.
By theorem 2.20 , $|\psi_{AB}⟩ = \sum_{i=1}^r \tilde s_i|\tilde e_i⟩ ⊗ |f_i⟩$, $\tilde s_i >0$ with the $|\tilde e_i\rangle\in H_A$ orthonormal and $|f_i\rangle \in H_B$ so that
$\rho_A=Tr_{B}(|\psi_{AB}\rangle\langle \psi_{AB}|)=\sum_{i=1}^r\tilde s_i^2|\tilde e_i\rangle \langle\tilde e_i|$
I can relate the bases $B=\{|e_i\rangle\}_{i=1}^n$ and $\tilde B=\{ |\tilde e_i\rangle\}_{i=1}^n$ by a change-of-basis matrix C, so that
$|\tilde e_i\rangle=\sum_{j=1}^r c_{j,i}|\tilde e_j\rangle$
$\sum_{i=1}^r \tilde s_i^2 |e_i\rangle\langle e_i|=\sum_{i=1}^r\sum_{j=1}^r\sum_{j'=1}^r \tilde s_i^2 c_{j,i}c_{j',i}|e_j\rangle\langle e_{j'}|$
But I cannot symplify this to get the desired form. What's going on? What should have been the correct approach?
You correctly deduced that singular vectors $\left|\tilde{e}_i\right>$ of $\Psi_{AB}$ will be eigenvectors of $\rho$. What is also important is that the eigenvalue of $\left|\tilde{e}_i\right>$ is equal to the square of the singular value $\tilde{s}_i$.
This implies that the change of basis from $B$ to $\tilde{B}$ is not arbitrary: $\left|\tilde{e}_i\right>$ is not just a linear combination of all eigenvectors $\left|e_i\right>$, it is a linear combination of only eigenvectors $\left|e_i\right>$ corresponding to the singular value $\tilde{s}_i^2$.
There are different ways to express this, it seems that you prefer to work with explicit sums, so I will also write in this form. Assume $$\tilde{s}_1 = \tilde{s}_2 = \dots = \tilde{s}_{q_1} > \tilde{s}_{q_1 + 1} = \dots = \tilde{s}_{q_2} > ... > \tilde{s}_{q_{t - 1} + 1} = \dots = \tilde{s}_{q_t}\quad\text{with $q_t = n$}$$ That is, there is only $t$ different singular values, and the $k$-th singular value appears with multiplicity $q_k - q_{k - 1}$, where I take $q_0 = 0$ to simplify the notation.
We have $$\left|\psi_{AB}\right> = \sum_{i = 1}^n \tilde{s}_i \left|\tilde{e}_i\right>\left|f_i\right> = \sum_{k = 1}^t \sum_{i = q_{k - 1} + 1}^{q_k} \tilde{s}_i \left|\tilde{e}_i\right>\left|f_i\right> = \sum_{k = 1}^t \sum_{i = q_{k - 1} + 1}^{q_k} \tilde{s}_{q_k} \left|\tilde{e}_i\right>\left|f_i\right>$$ and consequently $$\rho = \operatorname{Tr}_B\left|\psi_{AB}\rangle\langle \psi_{AB}\right| = \sum_{k = 1}^t \sum_{i = q_{k - 1} + 1}^{q_k} \tilde{s}_{q_k}^2 \left|\tilde{e}_i\rangle\langle\tilde{e}_i\right|$$ This means that:
Now if I order the eigenvectors $\left|e_i\right>$ in the same way, so that $\left|e_{q_{k - 1} + 1}\right>, \dots, \left|e_{q_k}\right>$ are eigenvectors corresponding to the eigenvalue $\tilde{s}_{q_k}^2$, then $\left|\tilde{e}_{q_{k - 1} + 1}\right>, \dots, \left|\tilde{e}_{q_k}\right>$ are linear combinations of $\left|e_{q_{k - 1} + 1}\right>, \dots, \left|e_{q_k}\right>$ and they are orthonormal bases of the same subspace — the $\tilde{s}_{q_k}^2$-eigenspace of $\rho$. It follows that $$\sum_{i = q_{k - 1} + 1}^{q_k} \left|\tilde{e}_i\rangle\langle\tilde{e}_i\right| = \sum_{i = q_{k - 1} + 1}^{q_k} \left|e_i\rangle\langle e_i\right|$$ and $$\rho = \sum_{k = 1}^t \sum_{i = q_{k - 1} + 1}^{q_k} \tilde{s}_{q_k}^2 \left|\tilde{e}_i\rangle\langle\tilde{e}_i\right| = \sum_{k = 1}^t \sum_{i = q_{k - 1} + 1}^{q_k} \tilde{s}_{q_k}^2 \left|e_i\rangle\langle e_i\right| = \sum_{i = 1}^n s_i^2\left|e_i\rangle\langle e_i\right|$$