If $a_1,...,a_m$ are rows of a matrix $A\in \Bbb R^{m\times n}$ do we have that $\langle a_i,a_1+...+a_m\rangle\geq0$ ?
For context I'm trying to prove that in a full rank bounded polyhedron $P\subset\Bbb R^n$, $P=\{x:Ax\leq b\}$ There exist vertices $v_1,...,v_{n+1}$ affinely independent.
The part I need this specific result for is showing that if $m>n$ (obviousely $m\geq n$ because of the full rank but the boundedness will imply $m>n$)
My reasoning:
If $m=n$ then by calling $x_0$ a basic feasible solution of $P$, the set $\{x_0-\lambda(a_1+...+a_n)\ :\ \lambda>0\}$ is an unbounded subset of $P$
In fact let $\lambda$ be positive. $A(x_0-\lambda(a_1+...a_n))\leq b-\lambda A(a_1+...+a_n)$ which I want to be smaller a equal to the vector $b$.
We have $$\langle a_1 + \dots +a_m, a_1 + \dots + a_m\rangle = \|a_1 + \dots + a_m\|^2 \ge 0$$ so by linearity $$\langle a_1, a_1 + \dots + a_m\rangle + \dots + \langle a_m, a_1 + \dots + a_m\rangle \ge 0$$ and therefore at least one of the terms $\langle a_i, a_1 + \dots + a_m\rangle$ must be nonnegative.