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Sorry my question was not clearly stated. I will ask it more clearly.

Let $G$ be a matrix with only nonnegative elements with linearly independent columns. Then there exists a column, ${\bf g}$ of $G$ such that the orthogonal projection of ${\bf g}$ on the remaining columns of G is a nonnegative linear combination of those columns. In other words, for a suitable column ${\bf g}$ of $G$ the vector ${\bf x}$ that minimizes $\parallel G^* {\bf x} - {\bf g}\parallel$ has nonnegative elements, where $G^*$ is a matrix consisting of all the columns of $G$ except ${\bf g}$.

Is the above a correct statement?

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I assume that by "orthogonal projection" you mean orthogonal projection to the span of the remaining column-vectors. The answer is negative if you have at least 4 column vectors. I will consider the case of 4 vectors, since the general case is similar. Assume first that you have four vectors in $C=(R_+)^3$, the non-negative octant in $R^3$. Take a homogeneous closed convex cone $Q\subset C$ which has four extremal rays $r_i, i=1,...,4$ generated by vectors $v_i$, $i=1,...,4$. Then for every $i$, $v_i$ does not belong to the nonnegative cone generated by the remaining vectors. In order to get four linearly independent vectors, just perturb $v_4$ a bit in $(R_+)^4$ and consider $w_1,...,w_4$, the resulting quadruple of vectors. Then projecting the nonnegative cone generated by the new vectors to the span of any of the 3 vectors would still have exactly 4 extremal rays, so you get the example.

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