Is idempotency, $rho^2=rho$, a necessary and sufficient condition for $rho$ to be a pure state?

THEOREM 1 If $rho$ is a density matrix (i.e., a positive, unit-trace, trace-class operator, also in an infinite dimensional Hilbert space), then $rho$ is a pure state iff $rho^2=rho$.

Proof. If $rho$ is pure, then $rho^2=rho$. Let us prove the converse implication. Suppose that $rho^2 = rho$ ($rho^2$ is trace class if $rho$ is because the set of trace class operators is a $^*$ ideal of the $C^*$-algebra of bounded operators) then $$0=rho^2-rho = sum_{j} (lambda_j^2 -lambda_j) |psi_jrangle langle psi_j|tag{1}$$ where, from the definition of density matrix (positive unit-trace trace-class operator) $$lambda_j in [0,1]tag{2}$$ and $$sum_j lambda_j =1tag{3}:.$$ Let us assume that there are at least two $jneq j'$ with $lambda_j,lambda_{j'}>0$. We conclude from (2) that both $lambda_j^2-lambda_j<0$ and $lambda_{j'}^2-lambda_{j'}<0$. Since $langle psi_k|psi_h rangle = delta_{hk}$, (1) leads to $$0 = langle psi_{j'}|0 psi_{j'}rangle = lambda_{j'}^2-lambda_{j'} <0$$ that is impossible. We conclude that the assumption that there are at least two $jneq j'$ with $lambda_j,lambda_{j'}>0$ is untenable so that $rho = |psi_jrangle langle psi_j|$. $Box$

With a similar route one easily proves that

THEOREM 2 If $rho$ is a density matrix (also in an infinite dimensional Hilbert space), then $rho$ is a pure state iff $tr(rho^2)=tr(rho)$ ($=1$).

Proof. If $rho$ is pure the thesis it trivial. Let us pass to the converse implication. Since $sum_j lambda_j =1$ and $lambda_jin [0,1]$, if more than one $lambda_j$ does not vanish, every $lambda_j$ is strictly less than $1$, so that we have in particular $lambda^2_j < lambda_j$ for all $j$, which implies $sum_j lambda_j^2 < sum_j lambda_j$. This meas that if $tr(rho^2) = tr(rho)$, then only one $lambda_j$ does not vanish so that $rho$ is pure. $Box$