({\ensuremath{|i\rangle}}{\ensuremath{\langlej|}})_A \otimes ({\ensuremath{|\mu\rangle}}{\ensuremath{\langle\nu|}})_B 
 = \sum_{k,l} \xi_{kl} \tilde{G}_{k}^{A} \otimes \tilde{G}_{l}^{B}.
\label{eq:ccn1}$$ We have used the same notations as in Eq. (\[eq:mixed\_state\]), except that we have added zeros to the tensor $a_{ij}^{\mu \nu}$, so that the indices run until the dimensions of the Hilbert spaces. $\{\tilde{G}_{k}^{A}\} = \{{\ensuremath{|i\rangle}}{\ensuremath{\langlej|}}\}$ and $\{\tilde{G}_{l}^{B}\} = \{{\ensuremath{|\mu\rangle}}{\ensuremath{\langle\nu|}}\} $ are complete sets of orthonormal Hermitian operators on the Hilbert spaces $\mathcal{H}_A$ and $\mathcal{H}_B$ respectively, with $ 1 \leq k \leq d_{A}^{2} $ and $ 1 \leq l \leq d_{B}^{2}$. Without loss of generality, we assume that $d_{A} \geq d_{B}$. After singular value decomposition of the matrix $\xi$, we have $$\xi = U \Sigma V^{\dagger},$$ where $U$ and $V$ are $ d_{A}^{2} \times d_{A}^{2} $ and $ d_{B}^{2} \times d_{B}^{2} $ dimensional unitary matrices respectively, and $ \Sigma $ is a $ d_{A}^{2} \times d_{B}^{2} $ dimensional diagonal matrix. Denoting the $k$th column vector of $ U $ and $ V $ by $ {\ensuremath{|u_k\rangle}} $ and $ {\ensuremath{|v_k\rangle}} $, the above expression becomes $$\xi = \sum_{k = 1}^{d_B^2} \lambda_{k} {\ensuremath{|u_{k}\rangle}} {\ensuremath{\langlev_{k}|}},$$ where $ \lambda_{k}$ are the diagonal elements of $\Sigma$. So, we have the matrix elements of $\xi$ as $$\xi_{kl} = \lambda_{k} \delta_{kl}.
  \label{eq:ccn2}$$ If $G_k^A$ and $G_l^B$ 