= Solution
Use the <Schmidt decomposition> of the bipartite <pure state>:
$$
|\psi\rangle_{QR}=\sum_{j=1}^r\sqrt{p_j}\,|u_j\rangle_Q|v_j\rangle_R,\qquad p_j>0,\quad\sum_jp_j=1.
$$
The two <reduced density matrices>, obtained by <partial trace>, are
$$
\rho_Q=\sum_jp_j|u_j\rangle\langle u_j|,\qquad\rho_R=\sum_jp_j|v_j\rangle\langle v_j|.
$$
Their nonzero <eigenvalues> are identical, even if the two ambient dimensions differ. Zero <eigenvalues> contribute no <Von Neumann entropy>, so
$$
\boxed{H(Q)_\psi=H(R)_\psi=-\sum_jp_j\log_2p_j.}
$$
This common <Von Neumann entropy> is the <entanglement entropy> of the bipartite <pure state>.
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