# Properties of vector combinations in the non-negative orthant

Given a vector $$x \in \mathbb{R}^{n}_{0+}$$ such that $$x = \sum^{k}_{i=1} \alpha_{i}v_{i}$$, the vectors $$(v_{1},...,v_{k}) \in \mathbb{R}^{n}_{0+}$$ are an independent set, $$k < n$$, and $$\alpha_{i} > 0$$, it can be seen by simple combinatorial argument that for at least one vector $$v_{i}$$, $$\frac{||x - \alpha_{i}v_{i}||_{2}}{||x||_{2}} \geq \frac{1}{k}$$ (assume the opposite, then we have that the maximum value of $$\frac{||x - \alpha_{i}v_{i}||_{2}}{||x||_{2}} < \frac{1}{k}$$ but this implies that $$||x - \sum^{k}_{i=1} \alpha_{i}v_{i}||_{2} > 0$$ which is a contradiction).

Say we have an additional vector $$y \in \mathbb{R}^{n}_{0+}$$ such that $$(v_{1},...,v_{k},y)$$ is an independent set. What can be said about $$\frac{||x - \beta y||_{2}}{||x||_{2}}$$ where $$\beta$$ is the result of minimizing $$||x-y\beta||_{2}$$ subject to $$\beta \geq 0$$? If $$(v_{1},...,v_{k},y)$$ is orthogonal then $$\frac{||x - \beta y||_{2}}{||x||_{2}} = 1$$, but bounds in the absence of orthogonality don't seem as obvious. If we impose the additional requirement that $$x - y\beta \geq 0$$ element-wise are there stronger conclusions that can be drawn?

P.S. Apologies if the title I chose for this question is not perfectly representative of what I actually asked, I was unable to determine if there is a common technical term for this sort of question.

EDIT: Removed silly equivocation about the case when $$(v_{1},...,v_{k},y)$$ is orthogonal.

EDIT 2: Further research revealed a flaw in the reasoning presented here. I'm in the process of reformulating the question and examining if I can come to any conclusions myself in the meantime. I may repost at a later date if I can't make further progress, but as written here this question is ill-posed.

• I don't get your point concerning orthogonality. If $(v_1,\ldots,v_k,y)$ is orthogonal, then $x$ is orthogonal to $y$. This implies $\beta = 0$. Thus, $\|x -\beta\,y\|/\|x\| = 1$? – gerw Feb 6 '19 at 12:47
• Yes, I was equivocating when I wrote the post in case I missed something stupid (and in the process wrote something vacuously true). I'll edit that section accordingly. – nick.schachter Feb 6 '19 at 15:08
• You have written "=0", but it should be "=1"? – gerw Feb 6 '19 at 15:19
• Just fixed it, thanks for the heads up. – nick.schachter Feb 6 '19 at 15:20