Geometric interpretation of 1-distillability

^{This is a sequel to Motivation for the definition of k-distillability}

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Geometrical interpretation from the definition of 1-distillability

• The eigenstate $|psirangle$ of the partially transposed $1$-distillable states will have Schmidt rank at most 2, i.e.,

$$|psiranglelanglepsi|=sum_ilambda_i^2|iiranglelangle ii|, textit{ where } sum_ilambda_i^2=1tag{17}$$

• The constraint $sum_ilambda_i^2=1$ gives rise to a geometric structure in arbitrary $N$ dimensions.

^Source

Questions:

1. In the definition of $k$-distillability (cf. here) we were talking about bipartite density matrices $rho$ in $H_Aotimes H_B$. In what sense is $|psirangle$ an "eigenstate" of a partially transposed $1$-distillable state? Is for $1$-distillable $rho$‘s the (non-normalized) state $|psirangle in H$, such that $langle psi|sigma|psirangle < 0$ necessarily an eigenstate of $rho$? Also, can we prove that $langle psi|sigma|psirangle < 0$ for any eigenstate of $rho$?

2. I do not see how the fact that "the eigenstate $|psirangle$ of the partially transposed $1$-distillable states will have a Schmidt rank at most 2" is encapsulated within the statement "$|psiranglelanglepsi|=sum_ilambda_i^2|iiranglelangle ii|$ where $sum_ilambda_i^2=1$".

   As far as I understand, the Schmidt decomposition of a pure state $|Psirangle$ of a composite system AB, considering an orthonormal basis $|i_Arangle$ for system A and $|i_Brangle$ for system B, is $$|Psirangle = sum_i lambda_i|i_Arangle|i_Brangle,$$ where $lambda_i$ are non-negative real numbers satisfying $sum_ilambda_i^2=1$ known as Schmidt co-efficients. Now the number of non-zero $lambda_i$‘s in the Schmidt decomposition is called Schmidt rank or Schmidt number. So I don’t quite understand the geometric constraint they’re talking about; if the Schmidt rank is at most 2, then we’d be restricted to only two cases i.e. $lambda_1^2=1$ and $lambda_1^2+lambda_2^2 = 1$…which aren’t so interesting. Am I missing something?