= Solution
For any normalized <pure state> on finite-dimensional <Hilbert spaces> $\mathcal H_A\otimes\mathcal H_B$, the <Schmidt decomposition> gives
$$
\boxed{|\psi\rangle_{AB}=\sum_{j=1}^r\sqrt{p_j}\,|u_j\rangle_A|v_j\rangle_B,\qquad
p_j>0,\quad\sum_jp_j=1,\quad r\leq\min(d_A,d_B).}
$$
The $u_j$ and $v_j$ form <orthonormal sets> in their respective spaces. The positive numbers $\sqrt{p_j}$ are the <Schmidt coefficients>, and $r$ is the <Schmidt rank>. Taking a <partial trace> gives the <reduced density matrices>
$$
\rho_A=\sum_jp_j|u_j\rangle\langle u_j|,\qquad
\rho_B=\sum_jp_j|v_j\rangle\langle v_j|.
$$
Thus the nonzero <eigenvalues> of the two <reduced density matrices> agree.
One construction is a <singular value decomposition> of the coefficient <matrix> $M$ in $|\psi\rangle=\sum_{a,b}M_{ab}|a\rangle|b\rangle$. If $M=U\Sigma V^\dagger$, the columns of $U$ give $u_j$, the complex conjugates of the columns of $V$ give $v_j$, and the <singular values> give $\sqrt{p_j}$. This also explains why the <Schmidt coefficients> are real and nonnegative.
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