I am interested in the linear least square problem with the solution constrained to the closed standard simplex:
$$ \min_x \|Ax-b\|^2$$ subject to $x_i \ge 0$ and $\| x \|_1 = 1$.
More precisely, I want to know which conditions on $A$ provides a unique solution to the problem.
I found conditions on the rank of $A$ for equality constraints from chapter 5 of Numerical methods for least square problems by Ake Bjorck, but I don't know if they apply to my case.
On
Start the linear system $$ \mathbf{A} x = b, $$ where $$ \mathbf{A}\in\mathbb{C}^{m\times n}, \ b\in\mathbb{C}^{m}, \ x\in\mathbb{C}^{n}. $$ The least squares solution always exists and, in most general form, is given by $$ x_{LS} = \mathbf{A}^{\dagger}b + \left( \mathbf{I}_{n} + \mathbf{A}^{\dagger}\mathbf{A}\right) y $$ where the arbitrary vector $y\in\mathbb{C}^{n}$ and $\mathbf{A}^{\dagger}$ is the Moore-Penrose pseudoinverse.
When $\mathbf{A}$ has full column rank (number of columns $n$ is the same as the matrix rank $\rho$), then the null space $\mathcal{N}\left( \mathbf{A} \right)$ is trivial and $$ \mathbf{A}^{\dagger}\mathbf{A} = \mathbf{I}_{n}, $$ and the least squares solution is a point. That is, the solution is unique.
Since $x_i \ge 0$, $\|x\|_1 = \sum_i x_i$, and the feasible region is convex (in fact a simplex). If $A$ is injective on $L = \{u: \sum_i u_i = 0\}$, the objective is strictly convex so the minimizer is unique. On the other hand, in order for the minimizer $p$ to be unique there can't be any nonzero $u \in L$ with $Au = 0$ and $p+u$ feasible. Thus if $u \in L$ and $Au = 0$ there must be some index $i$ such that $p_i = 0$ and $u_i < 0$ (so that $p + t u$ is infeasible for all $t > 0$) and some $i$ such that $p_i = 0$ and $u_i > 0$ (so that $p + t u$ is infeasible for all $t < 0$).