= Solution
The coefficient <matrix> in the computational <orthonormal bases>, with Alice's index labelling rows, is
$$
C=\frac1{\sqrt3}\begin{pmatrix}0&-1\\1&1\end{pmatrix}.
$$
Since the entries are real, the <reduced density matrices> are
$$
\rho_A=CC^\dagger=\frac13\begin{pmatrix}1&-1\\-1&2\end{pmatrix},
\qquad
\rho_B=C^{\mathsf T}\overline C=\frac13\begin{pmatrix}1&1\\1&2\end{pmatrix}.
$$
Both have <trace> one and <determinant> $1/9$. Their characteristic <polynomial> is $\lambda^2-\lambda+1/9$, so the nonzero <eigenvalues> are
$$
\lambda_\pm=\frac{3\pm\sqrt5}{6}.
$$
Both are positive. Thus the <Schmidt rank>, or the <Schmidt number> of this <pure state>, is
$$
\boxed{r=2.}
$$
If “Schmidt numbers” denotes the individual positive <Schmidt coefficients> instead, those are
$$
\boxed{s_+=\frac{\sqrt5+1}{2\sqrt3},\qquad s_-=\frac{\sqrt5-1}{2\sqrt3}.}
$$
Their squares are $\lambda_\pm$; giving both the <Schmidt rank> and the <Schmidt coefficients> removes this terminology ambiguity. The <pure state> is <entangled> because its <Schmidt rank> exceeds one.
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