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Choose an optimal feasible vector $x$ having as few positive coordinates as possible, and letIf the columns $(A_i{)}_{i\in S}$ were linearly dependent, there would be a nonzero vector $h$, supported on $S$, such that $Ah=0$. For all sufficiently small positive $\epsilon$, bothwould remain feasible.

If $c^{T}h\ne 0$, one of these two perturbations would increase the objective, contradicting optimality. Hence $c^{T}h=0$. We may then increase $\epsilon$ in one of the two directions until at least one positive coordinate first becomes zero. The resulting vector is still feasible and optimal but has smaller positive support, contradicting the choice of $x$.

Let $(y,t)$ be a basic feasible solution of $R$. Part (ii) gives $t>0$, and put $x=y/t$. We already know that $x$ is feasible for $P$.

Suppose the columns $A_i$ with $y_i>0$ were linearly dependent. Then some nonzero $h$, supported on those indices, would satisfy $Ah=0$. PutSince $d^{T}y=1$ and $Ay=bt$,The vector $(\delta y,\delta t)$ is nonzero and is supported only on positive variables of $(y,t)$. It is therefore a linear dependence among the active constraint columns of $R$, contradicting that $(y,t)$ is basic. Hence the active columns $A_i$ are linearly independent, and

By the Fundamental theorem of linear programming, $R$ has an optimal basic feasible solution. Its image $x\in \mathcal{B}$ has the same objective ratio, by part (ii). Conversely, every $x\in \mathcal{B}$ is feasible for $Q$ and maps to a feasible point of $R$. Since $x^{\ast }$ solves $Q$,