= Solution
Diagonalize the <reduced density matrix> on Alice's <Hilbert space>:
$$
\rho_A=\sum_{j=1}^r\lambda_j|a_j\rangle\langle a_j|,
\qquad\lambda_j>0.
$$
For every positive <eigenvalue>, define a vector on Bob's <Hilbert space> by
$$
|b_j\rangle=\lambda_j^{-1/2}(\langle a_j|\otimes I_B)|\Psi_{AB}\rangle.
$$
The definition of the <partial trace> gives
$$
\langle b_i|b_j\rangle
=\frac{\langle a_j|\rho_A|a_i\rangle}{\sqrt{\lambda_i\lambda_j}}
=\delta_{ij}.
$$
Thus the $|b_j\rangle$ are <orthonormal>. Extend the $|a_j\rangle$ to an <orthonormal basis> of Alice's <Hilbert space>. If $|a\rangle$ lies in the zero-<eigenvalue> subspace of $\rho_A$, then
$$
\| (\langle a|\otimes I_B)|\Psi_{AB}\rangle\|^2
=\langle a|\rho_A|a\rangle=0.
$$
There is therefore no component of $|\Psi_{AB}\rangle$ in those additional basis directions. Expansion in Alice's <orthonormal basis> yields the <Schmidt decomposition>
$$
\boxed{|\Psi_{AB}\rangle=\sum_{j=1}^r\sqrt{\lambda_j}|a_j\rangle|b_j\rangle,
\quad\sum_j\lambda_j=1.}
$$
Its positive <Schmidt coefficients> are $s_j=\sqrt{\lambda_j}$ and its <Schmidt rank> is $r$. Taking the two <partial traces> and using the <orthonormality> of both families gives
$$
\rho_A=\sum_j\lambda_j|a_j\rangle\langle a_j|,
\qquad
\rho_B=\sum_j\lambda_j|b_j\rangle\langle b_j|.
$$
Hence \b[the two reduced density matrices have exactly the same nonzero eigenvalues, including multiplicities]. Their numbers of zero <eigenvalues> may differ when the local <Hilbert spaces> have different dimensions. In particular their <Von Neumann entropies> agree.
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