= Solution
Take the vectors in the <quantum state ensemble> to be normalized, with $p_i\geq0$ and $\sum_i p_i=1$. The corresponding <density matrix> is
$$
\boxed{\rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|.}
$$
Each rank-one projector is a <self-adjoint operator>, so $\rho^\dagger=\rho$. For every $|v\rangle$ in the <Hilbert space>,
$$
\langle v|\rho|v\rangle=\sum_i p_i|\langle\psi_i|v\rangle|^2\geq0,
$$
proving that $\rho$ is <positive semidefinite>. The <trace> is $\operatorname{Tr}\rho=\sum_i p_i\langle\psi_i|\psi_i\rangle=1$.
Conversely, diagonalize a <self-adjoint operator> $\rho$ in an <orthonormal eigenbasis>, writing $\rho=\sum_j\lambda_j|e_j\rangle\langle e_j|$. Its <positive semidefiniteness> implies $\lambda_j=\langle e_j|\rho|e_j\rangle\geq0$, and its unit <trace> gives $\sum_j\lambda_j=1$. Thus $\{\lambda_j,|e_j\rangle\}$ is a <quantum state ensemble> with precisely this <density matrix>. Terms with zero <eigenvalue> may be omitted. This argument includes rank-one <pure states>; an ensemble representation need not describe a genuinely <mixed state>.
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