= Solution
A <density operator> is a positive self-adjoint <trace-class operator> with $\boxed{\operatorname{Tr}\rho=1}$. A finite <expectation> is $\boxed{\langle A\rangle_\rho=\operatorname{Tr}(\rho A)}$. It represents a <pure state> exactly when $\rho=|v\rangle\langle v|$ for a unit vector, equivalently $\rho^2=\rho$ or $\operatorname{Tr}\rho^2=1$.
For a <finite-dimensional vector space>[finite-dimensional] <spectral decomposition> $\rho=\sum_jp_j|v_j\rangle\langle v_j|$, introduce an auxiliary space with <orthonormal> vectors $f_j$ and use the enlarged space $\mathcal K=\mathcal H\otimes\mathcal E$. Define
$$
\Omega_\rho=\sum_j\sqrt{p_j}\,v_j\otimes f_j,\qquad\widetilde A=A\otimes I.
$$
Then $\|\Omega_\rho\|^2=\sum p_j=1$ and $\langle\Omega_\rho,\widetilde A\Omega_\rho\rangle=\sum p_j\langle v_j,Av_j\rangle=\operatorname{Tr}(\rho A)$. This <purification> shows that subsystem mixedness can arise from <entanglement> with an unobserved environment <even> when the total state is pure. Subsystem measurements alone cannot distinguish these two descriptions.
A nonzero pure bipartite vector is entangled when it cannot be written as a <tensor product> of two vectors; equivalently its <Schmidt rank> exceeds one. For mixed bipartite states, <entanglement> means absence of a convex decomposition into product <density operators>.
For the nonorthogonal expansion in the question, the unnormalized reduced operator is
$$
\boxed{S_1=\sum_{j,k}\langle\eta_k,\eta_j\rangle\,|\xi_j\rangle\langle\xi_k|.}
$$
Indeed $\langle\Psi,(A\otimes I)\Psi\rangle=\sum_{j,k}\langle\xi_k,A\xi_j\rangle\langle\eta_k,\eta_j\rangle=\operatorname{Tr}(AS_1)$. Positivity follows because $\langle v,S_1v\rangle=\|\sum_j\langle v,\xi_j\rangle\eta_j\|^2\geq0$, and $\operatorname{Tr}S_1=\|\Psi\|^2=:Z$. Thus $\boxed{\rho_1=S_1/Z}$ is a <density operator>; if $\Psi$ is already <normalization>[normalized], $Z=1$. The Gram factors cannot be dropped merely because the vectors are called bases. For unnormalized $\Psi$, physical <expectations> likewise divide its raw quadratic form by $Z$.
For the <Schmidt decomposition> diagonalize $S_1$ in an <orthonormal> <basis> $\xi_j$ with its positive <eigenvalues> $c_j^2$. Expand $\Psi=\sum_j\xi_j\otimes w_j$ with $w_j=(\langle\xi_j|\otimes I)\Psi$. The reduced <matrix> gives $\langle w_k,w_j\rangle=c_j^2\delta_{jk}$. Therefore $\eta_j=w_j/c_j$ are <orthonormal> and
$$
\boxed{\Psi=\sum_{j:c_j>0}c_j\xi_j\otimes\eta_j.}
$$
Zero-<eigenvalue> components vanish because their squared <norms> are zero. This proves the decomposition for an arbitrary nonzero vector; the zero vector uses the empty sum. For a unit vector $\sum c_j^2=1$.
In the numerical example the coefficient <matrix> in the two product bases is $C=\begin{pmatrix}1&2&3\\0&1&2\end{pmatrix}$. Taking the first-subsystem <trace> yields
$$
\boxed{S_2=C^{\mathsf T}C=\begin{pmatrix}1&2&3\\2&5&8\\3&8&13\end{pmatrix}.}
$$
The vector $(1,-2,1)^{\mathsf T}$ is in its <nullspace>, so <rank> is at most two; its leading $2\times2$ minor has <determinant> one, so $\boxed{\operatorname{rank}S_2=2}$. This is the <rank> of $C$ and the <Schmidt rank>, proving <entanglement>. \b[The displayed source ket has <norm> squared $19$], and the displayed reduced <matrix> has <trace> $19$. The physical <density operator> is $\boxed{\rho_2=S_2/19}$; its nonzero <eigenvalues> are $(19\pm\sqrt{337})/38$, since $CC^{\mathsf T}=\begin{pmatrix}14&8\\8&5\end{pmatrix}$. The raw <matrix> is correct for the unnormalized quadratic form actually printed.
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