 Solution

The Schmidt decomposition states that every finite-dimensional bipartite pure state haswhere $s_j>0$, ${\sum }_j{s}_j^2=1$, and the two displayed families are orthonormal. The integer $r$ is the Schmidt rank.

To prove it, choose product bases and writeApply the singular value decomposition $C=US{V}^{†}$. Absorbing the columns of $U$ and the complex conjugates of the columns of $V$ into new orthonormal bases gives the stated sum, with the nonzero singular values as the Schmidt coefficients. Equivalently, $s_j^2$ are the common nonzero eigenvalues of the two reduced density matrices, so $r=\mathrm{rank}{\rho }_A=\mathrm{rank}{\rho }_B$.