\end{align}where s_j(A)\ge 0 are the (always real and non-negative) singular values of A, as in the singular value decomposition. The inequality is saturated and becomes an equality when \langle w_j|a_j\rangle=1, that is, when U=\sum_k |b_k\rangle\!\langle a_k|, and thus AU=\sqrt{AA^\dagger}\equiv |A|. The above shows that |\langle\psi_\rho|\psi_\sigma\rangle|=
