Let $A \in \mathbb R^{m \times n}$ have rank $m < n$
On the way to showing that the condition (for $x \in \mathbb R^n$)
- $A x = b$
- $x \geq 0$
is equivalent to (intersection of halfspaces)
- $\forall i \in \lbrace n - m+1, ..., n \rbrace,~ b_i - \sum_{j=1}^{n-m} A^i_jx^j\geq 0$
- $\forall i \in \lbrace 1, ..., n-m \rbrace,~ x^i \geq 0$
these authors introduce a little hack (chapter 2 section 2.3.3):
$\forall i \in \lbrace n-m+1, ..., n\rbrace ,~ x_i = b_i - \sum_{j=1}^{n-m} A^i_j x^j$
They reason that this is because we can pre-multiply the orginal form by a basis $B$ for $A$ which is (WLOG) constructed of the last $m$ columns, which should directly yield this.
So, declaring the subsequence $\sigma : [m] \rightarrow [n]$ where $\sigma k = k + n - m$, we can write our basis matrix as $B_k = A_{\sigma k}$, which we (may) assume is invertible.
It follows that $\sum_j (B^{-1})^i_j A^j_{\sigma k} = \sum_j (B^{-1})^i_j B^j_k = I^i_k$
Then pre-multiplying $Ax = b$ with $B^{-1}$ as suggested becomes
$(B^{-1} b)^i = (B^{-1} A x)^i$
$ = \sum_{j=1}^m \sum_{l=1}^n (B^{-1})^i_j A^j_l x^l$
$ = \sum_{j=1}^m \sum_{l=1}^{n-m} (B^{-1})^i_j A^j_l x^l + \sum_{k=1}^{m} \sum_{j=1}^m (B^{-1})^i_j A^j_{\sigma k} x^{\sigma k}$
$ = \sum_{j=1}^m \sum_{l=1}^{n-m} (B^{-1})^i_j A^j_l x^l + \sum_{k=1}^{m} I^i_k x^{\sigma k}$
$ = \sum_{j=1}^m \sum_{l=1}^{n-m} (B^{-1})^i_j A^j_l x^l + x^{\sigma i}$
So I get
$x^{i+n-m} = (B^{-1} (b - Cx))^i$
Where $C = [A_1 ~ ... ~ A_{n-m} ~ 0 ~ ... ~ 0]$ where the last $m$ columns are zero
How do we complete the hack? Is their form (and reasoning) correct?