I'm having difficulty following a proof in the book Nonlinear Programming Theory and Algorithms, Third Edition by Bazaraa, Sherali, and Shetty on page 68 section 2.6.4. The proof is reproduced below, up to the part of the proof I'm confused about.
My question concerns the author's statement:
But this implies that $x_{11} = x_{21} = B^{-1}b$.
I don't see how the author makes that connection.
Theorem (Existence of Extreme Points)
Let $S = \{ x : Ax = b, x \geq 0\}$, where $A$ is an m x n matrix and $b$ is an m-vector. A point $x$ is an extreme point of $S$ if and only if $A$ can be decomposed into $[B, N]$ such that
$x = \begin{bmatrix} x_B \\ x_N \end{bmatrix} = \begin{bmatrix} B^{-1}b \\ 0 \end{bmatrix}$
Proof:
Suppose that $A$ can be decomposed into $[B, N]$ with $x = \begin{bmatrix} B^{-1}b \\ 0 \end{bmatrix}$ and $B^{-1}b \geq 0$. It is obvious that $x \in S$. Now suppose that $x = \lambda x_1 + (1 - \lambda)x_2$ with $x_1$, $x_2 \in S$ for some $\lambda \in (0,1)$. In particular, let $x^{t}_{1} = (x^{t}_{11}, x^{t}_{12})$ and $x^{t}_{2} = (x^{t}_{21}, x^{t}_{22})$. Then
$\begin{bmatrix} B^{-1}b \\ 0 \end{bmatrix} = \lambda \begin{bmatrix} x_{11} \\ x_{12} \end{bmatrix} + (1 - \lambda) \begin{bmatrix} x_{21} \\ x_{22} \end{bmatrix}$
Since $x_{12}, x_{22} \geq 0$ and $\lambda \in (0,1)$, it follows that $x_{12} = x_{22} = 0$. But this implies that $x_{11} = x_{21} = B^{-1}b$ and hence $x = x_1 = x_2$. This shows that $x$ is an extreme point of $S$.