Prove $P = \{x | Ax = b, x ≥ 0 \}$ has at most two BFSs if $A$ has dimensions $m\times (m+1)$ Consider the standard form polyhedron $P = \{x | Ax = b, x ≥ 
0 \}$. Suppose that the matrix $A$ has dimensions $m \times n$ and that its rows are linearly independent. Prove that if $n = m + 1$, then $P$ has at most two basic feasible solutions.
I know this question has already been asked here, but I didn't understand the solution -- like at all. If possible, can someone please share an alternative approach? Thanks..
 A: I'll attempt to clear up some confusion with the approach you linked. First notice that your polyhedron $P$ is a hyperplane, since the polyhedron is written in standard form.
Given that the number of variables in your problem is one more than the number of constraints $(n=m+1)$ and that the constraints are linear independant, we know from linear algebra that the nullspace of $A$ is one-dimensional.
Let us assume that $P$ is non-empty, i.e. $\exists y \in \mathbb{R}^n : y\geq 0 \text{ and } Ay=b$, we know that taking any non-zero $z \in \text{null }A$ we can create another solution $y+\lambda z,\; \forall \lambda \in \mathbb{R}$ that satisfies the main constraints, since $A(y+\lambda z)=Ay+\lambda Az=Ay = b$.
Since $\text{null A}$ is a 1-dimensional subspace of $\mathbb{R^n}$ we know that the solution space $S$ of the linear equation of the form $Ax=b$ is also 1-dimensional thus any element in $P$ can be written in the form $y+\lambda z$ for suitable real $\lambda$. Any element of $P$ will also be an element of $S$. Thus $P$ is a subset of a line in $\mathbb{R}^n$. Notice that the possible convex subsets of a line are the line itself, a half-line and a line-segment, this will give you a geometric understanding of the proof.
Assume by contradiction that $P$ has 3 distinct basic feasible solutions $x_1, x_2, x_3 \in P$. Since they all reside on the line, one of them must be between the two others. WLOG assume $x_3$ is that extreme point. Since it is between the two other solutions and $P$ is convex, we can write $x_3$ as a convex combination of $x_1$ and $x_2$. Thus $x_3$ is not an extreme point and neither a basic feasible solution, which contradicts our assumption.
One can find an explicit way of computing the convex combination to create $x_3$, by letting one of the basic feasible solutions "span" the affine subspace that is $S$
