= Solution
A <density operator> is a positive self-adjoint trace-class operator with trace one. In finite dimensions this is a positive semidefinite <Hermitian matrix> with $\operatorname{tr}\rho=1$. Its expectation of an observable is $\boxed{\langle A\rangle=\operatorname{tr}(A\rho)}$ whenever the expectation is defined.
In the source's square-root notation, take the unique positive square root $W=\sqrt\rho$. It has Hilbert-Schmidt <norm> one and satisfies $WW^*=\rho$, so
$$
\boxed{\operatorname{tr}(A\rho)=\operatorname{tr}(W^*AW).}
$$
The standard vector <purification> is the associated vector $|W\rangle\!\rangle=\sum_{r,s}W_{rs}|\xi_r\rangle|\eta_s\rangle$ on a system and a chosen copy. Tracing out the copy gives $\rho$; measuring $A\otimes I$ gives the same trace formula. Thus the operator square root and the purified vector are two related objects, distinguished by this vectorization.
For a general coefficient matrix $C$, expansion of the expectation gives
$$
\langle\Psi|A\otimes I|\Psi\rangle
=\sum_{r,t,s}\overline{c_{rs}}A_{rt}c_{ts}
=\operatorname{tr}(ACC^*).
$$
Thus $\boxed{\rho_1=CC^*}$. Positivity is immediate from $v^*CC^*v=\|C^*v\|^2$, and its trace is one by <normalization> of $\Psi$.
If $C$ is nonsingular, it is square and $\rho_1$ is positive definite. Define
$$
\boxed{U=C^{-1}\sqrt{\rho_1}.}
$$
Then $UU^*=C^{-1}\rho_1(C^*)^{-1}=I$, which for a square matrix also implies $U^*U=I$. Thus $U$ is unitary, $\sqrt{\rho_1}=CU$, and $C=\sqrt{\rho_1}U^*$.
In an eigenbasis of $\rho_1$, its square root is $\operatorname{diag}(\sqrt{p_r})$. Hence $c_{rs}=\sqrt{p_r}\,\overline{U_{sr}}$. Define
$$
\boxed{|\zeta_r\rangle=\sum_s\overline{U_{sr}}|\eta_s\rangle.}
$$
Unitarity gives $\langle\zeta_r|\zeta_t\rangle=\delta_{rt}$, and substitution gives the <Schmidt decomposition>
$$
\boxed{|\Psi\rangle=\sum_r\sqrt{p_r}|\xi_r\rangle|\zeta_r\rangle.}
$$
For singular or rectangular $C$, the same conclusion follows on its support by a <singular value decomposition>; the stated nonsingular case needs only the unitary calculation above.
For the supplied <Bell state>, the requested matrices are
$$
\boxed{C=\frac1{\sqrt2}\begin{pmatrix}0&1\\-1&0\end{pmatrix},\quad
\rho_1=\frac12I,\quad\sqrt{\rho_1}=\frac1{\sqrt2}I,\quad
U=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.}
$$
The Schmidt vectors can be chosen as $\zeta_1=\eta_2$, $\zeta_2=-\eta_1$, and the two coefficients are $1/\sqrt2$. The square root's canonical vector <purification> is $(|\xi_1\eta_1\rangle+|\xi_2\eta_2\rangle)/\sqrt2$; the original Bell vector is a different <purification> of the same reduced state, related by a unitary on the second factor. This illustrates the unitary freedom of <purification>.
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