Are all extremal points of the feasible set of an arbitrary affine equation pure states?

Question

Suppose I have one constraint on quantum states, i.e., $\Lambda(\rho)=Y$ where $Y$ is a Hermitian matrix and $\Lambda$ is a linear and Hermitian preserving map. Note that $\rho$ and $Y$ can in general have different dimensions. Let's denote all quantum states satisfy the constraint as the set $\mathcal S$. My question is, for an arbitrary linear and Hermitian preserving map $\Lambda$, are all the extreme points of $\mathcal S$ must be pure states? In other words, can we find one linear and Hermitian preserving map $\Lambda$ with one mixed state $\rho$ an extreme point of $\mathcal S$?

Edit

I've changed the term affine equality constraints concerning $\Lambda$ into $\Lambda$ is a linear and Hermitian preserving map. This type of constraint is a building block of Semidefinite programming(see, e.g., Watrous' book The Theory of Quantum Information).

Answer 1

A simple counterexample but perhaps I'm misinterpreting "one affine equation". Take the map $\Lambda$ to be the identity map and let $Y$ be any mixed state. Then the set of states satisfying the condition is a singleton set containing only a mixed state by construction.

Answer 2

This is generally not the case. One example is the set of CPTP maps. Under the Choi-Jamiołkowski isomorphism, this set corresponds to a set of quantum states intersected with an affine hyperplane given by the trace-preserving condition. A theorem by Choi characterizes the extreme points of this set (see e.g. Theorem 2.31 in Watrous' book), and they are not necessarily of rank one, i.e. not necessarily pure. An example is also given in Watrous' book.