= Solution
A normalized <pure state> has $\rho=|\psi\rangle\langle\psi|$, and therefore
$$
\boxed{\rho^2=\rho,\qquad\operatorname{Tr}\rho^2=1.}
$$
Conversely, diagonalize a permissible <density operator>. Its <eigenvalues> obey $0\le\lambda_j\le1$ and $\sum_j\lambda_j=1$, so
$$
1-\operatorname{Tr}\rho^2
=\sum_j\lambda_j(1-\lambda_j)\ge0.
$$
Equality requires every <eigenvalue> to be zero or one; exactly one is one. Thus the <idempotent matrix> condition and <purity> one both characterize a <pure state>. A <mixed state> has at least two positive <eigenvalues>, \b[$\rho^2\ne\rho$ and $\operatorname{Tr}\rho^2<1$]. In <dimension> $d$ its <purity> is at least $1/d$.
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