Purifications exist and are unitarily equivalent after padding

Use [F1] and let $K_0={\mathbb{C}}^r$ with the standard ordered basis relabelled $f_1,\dots ,f_r$, orthonormal by [F9]. Define ${\mathrm{\Psi }}_0={\sum }_{j=1}^r\sqrt{{\lambda }_j}{e}_j\otimes f_j$. The product pairing [F4] gives $\mathrm{\parallel }{\mathrm{\Psi }}_0{\mathrm{\parallel }}^2={\sum }_j{\lambda }_j=1$, and [F5] gives ${\mathrm{Tr}}_{K_0}(\mathrm{\mid }{\mathrm{\Psi }}_0⟩⟨{\mathrm{\Psi }}_0\mathrm{\mid })={\sum }_j{\lambda }_j\mathrm{\mid }{e}_j⟩⟨e_j\mathrm{\mid }=\rho$. Thus ${\mathrm{\Psi }}_0$ is a purification on an auxiliary space of dimension $r=\mathrm{rank}(\rho )$ by [F12].

For each $i$, choose an orthonormal basis $(g_j^{(i)}{)}_{j<d_i}$ of $K_i$ by [F6], and put $h_j^{(i)}={\iota }_i{g}_j^{(i)}$. Since ${\iota }_i$ is linear and norm-preserving by [F11], each $h_j^{(i)}$ has norm one. For $j\mathrm{\ne }\mathrm{\ell }$, the four vectors $g_j^{(i)}\pm g_{\mathrm{\ell }}^{(i)}$ and $g_j^{(i)}\pm i{g}_{\mathrm{\ell }}^{(i)}$ have squared norm $2$, so their images do too. With the root inner product convention, for $z=⟨h_j^{(i)},h_{\mathrm{\ell }}^{(i)}⟩$ the differences $\mathrm{\parallel }{h}_j^{(i)}+h_{\mathrm{\ell }}^{(i)}{\mathrm{\parallel }}^2-\mathrm{\parallel }{h}_j^{(i)}-h_{\mathrm{\ell }}^{(i)}{\mathrm{\parallel }}^2=2(z+\bar{z})$ and $\mathrm{\parallel }{h}_j^{(i)}+i{h}_{\mathrm{\ell }}^{(i)}{\mathrm{\parallel }}^2-\mathrm{\parallel }{h}_j^{(i)}-i{h}_{\mathrm{\ell }}^{(i)}{\mathrm{\parallel }}^2=2i(\bar{z}-z)$ both vanish. Thus $z+\bar{z}=0$ and $z=\bar{z}$, so $z=0$; the image vectors are orthonormal. Let $L_i=\mathrm{span}\{h_j^{(i)}:j<d_i\}$. By [F6], ${\mathbb{C}}^m=L_i\oplus L_i^{\perp }$ and $L_i^{\perp }$ is finite-dimensional; choose an orthonormal basis of $L_i^{\perp }$ by [F6] and append it to $(h_j^{(i)})$ to obtain an ordered orthonormal basis of ${\mathbb{C}}^m$. Choose an ordered orthonormal basis of $H$ by [F6]. The product basis result [F13] shows that $I_H\otimes {\iota }_i$ sends the orthonormal product basis of $H\otimes K_i$ to an orthonormal family, so the padded vector ${\tilde{\mathrm{\Psi }}}_i=(I_H\otimes {\iota }_i){\mathrm{\Psi }}_i$ is still a unit vector. Write ${\mathrm{\Psi }}_i={\sum }_{\alpha ,j}{c}_{\alpha j}{e}_{\alpha }\otimes g_j^{(i)}$ in this product basis. In the extended basis of ${\mathbb{C}}^m$, ${\tilde{\mathrm{\Psi }}}_i={\sum }_{\alpha ,j}{c}_{\alpha j}{e}_{\alpha }\otimes h_j^{(i)}$ has zero coefficients on every complement basis vector, so the partial-trace coordinate sum [F5] over the extended basis equals the original partial trace over $(g_j^{(i)})$, namely $\rho$. By [F3] write each ${\tilde{\mathrm{\Psi }}}_i$ in Schmidt form ${\sum }_j{s}_j^{(i)}{u}_j^{(i)}\otimes f_j^{(i)}$; [F5]–[F7] give its marginal ${\sum }_j(s_j^{(i)}{)}^2\mathrm{\mid }{u}_j^{(i)}⟩⟨u_j^{(i)}\mathrm{\mid }=\rho$. Therefore the vectors with $(s_j^{(i)}{)}^2=\lambda$ form an orthonormal basis of each positive $\lambda$-eigenspace of $\rho$.

Fix an eigenvalue $\lambda >0$. By step 1.2, its two lists of Schmidt vectors are orthonormal bases of the same eigenspace. Write $u_{\lambda ,j}^{(2)}={\sum }_k(V_{\lambda }{)}_{kj}{u}_{\lambda ,k}^{(1)}$. Orthonormality gives ${\sum }_k(V_{\lambda }{)}_{kj}\overline{(V_{\lambda }{)}_{k\mathrm{\ell }}}={\delta }_{j\mathrm{\ell }}$; conjugating this identity gives $V_{\lambda }^{\ast }{V}_{\lambda }=I$, so the square matrix $V_{\lambda }$ is unitary. Prescribe $f_{\lambda ,j}^{(2)}\mapsto {\sum }_k\overline{(V_{\lambda }{)}_{kj}}{f}_{\lambda ,k}^{(1)}$. For indices $j,\mathrm{\ell }$, the inner product of the two images is ${\sum }_k\overline{(V_{\lambda }{)}_{kj}}(V_{\lambda }{)}_{k\mathrm{\ell }}={\delta }_{j\mathrm{\ell }}$; distinct eigenvalue blocks remain orthogonal. Thus this basis map extends linearly by [F10] to an onto isometry between the Schmidt-support subspaces.

By [F6], each padded auxiliary space is the orthogonal direct sum of its Schmidt-support subspace and its orthogonal complement. Step 1.2 shows that each support has dimension $r$, so [F8] gives dimension $m-r$ for both complements. Choose orthonormal bases for them by [F6] and prescribe a bijection between those bases. The complement map and the support isometry from step 2.1 together send an orthonormal basis of ${\mathbb{C}}^m$ to an orthonormal basis of ${\mathbb{C}}^m$; [F10] extends this map uniquely and linearly, and the basis map is surjective and preserves norms. Hence [F11] makes it a unitary $U$. In each eigenvalue block, ${\sum }_j(V_{\lambda }{)}_{kj}\overline{(V_{\lambda }{)}_{\mathrm{\ell }j}}={\delta }_{k\mathrm{\ell }}$, and therefore ${\sum }_j\sqrt{\lambda }{u}_{\lambda ,j}^{(2)}\otimes U{f}_{\lambda ,j}^{(2)}={\sum }_{k=1}^{m_{\lambda }}\sqrt{\lambda }{u}_{\lambda ,k}^{(1)}\otimes f_{\lambda ,k}^{(1)}\mathrm{.}$ Summing these equalities over the positive eigenvalues gives the displayed equivalence after padding.