Let $P = \{ x\in \mathbb{R}^n | Ax = b, x\ge 0\}.$ Assume $rank(A)=m$, full rank.
- How do we show that dimension ($P$) $\le n-m$
- If $x>0$ how do we show that dimension $(P) = n-m$?
My idea may not make sense:
rank$(A)=\min\{m,n\}=m.$ Basis matrix $B,$ always has dimension $m \times m.$ Assume dim$(P) > n-m \Rightarrow$ dim$(B) > n-m$ which is a contradiction
Since rank(A) = $\min\{m,n\} = m$. Therefore $m\le n$. By rank-nullity theorem: rank$(A) +$ Nul$(A) = n$ we have that Nul$(A) = n - m.$ Since P has a non-degenerate solution $x>0$, then P has a unique solution (with a unique basis matrix) iff $Ax=b$ has a unique solution iff $Ax = 0$ has a unique solution, $x=0$, iff $x \in$ Ker$(A) =$ dim$(Nul(A))$. Thus $P$ has dimension $n-m$