= Solution
The coefficient <matrix> in the <computational basis> is
$$
M=\frac1{\sqrt{15}}\begin{pmatrix}2\sqrt2&-1\\\sqrt2&2\end{pmatrix}.
$$
For a bipartite <pure state>, the <partial traces> give $\psi_A=MM^\dagger$ and $\psi_B=M^T\overline M$. Therefore \b[the requested reduced density matrices are]
$$
\boxed{\psi_A=\frac1{15}\begin{pmatrix}9&2\\2&6\end{pmatrix},\qquad
\psi_B=\frac1{15}\begin{pmatrix}10&0\\0&5\end{pmatrix}.}
$$
The columns of $M$ are orthogonal. Their norms are $\sqrt{2/3}$ and $\sqrt{1/3}$, so normalizing them gives the <orthonormal set>
$$
|u_0\rangle=\frac{2|0\rangle+|1\rangle}{\sqrt5},\qquad
|u_1\rangle=\frac{-|0\rangle+2|1\rangle}{\sqrt5}.
$$
With $|v_0\rangle=|0\rangle$ and $|v_1\rangle=|1\rangle$, a <Schmidt decomposition> is
$$
\boxed{|\psi\rangle=\sqrt{\frac23}|u_0\rangle|0\rangle+
\sqrt{\frac13}|u_1\rangle|1\rangle.}
$$
Expanding this expression reproduces the minus sign in the $|01\rangle$ coefficient. The two <reduced density matrices> both have <eigenvalues> $2/3$ and $1/3$, consistent with these <Schmidt coefficients>.
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