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We suppose linear combination $a_1x_1+a_2x_2+\cdots+a_nx_n=y$, where $y>0$. If $x_i>0$, $i=1,2,\ldots n$, then $a_i$, $i=1,2,\ldots n$, have what conditions to satisfy the above equation? Conversely, What can be said about the $a_i$ to satisfy the above equation with assumption $x_i>0$?

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The only thing you can say is: At least one of $a_i$ must be positive.

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A system of vector basis is linear independent if no linear combination of them equals to zero i.e. let $x_1,x_2,...,x_n$ be a set of $n$ vectors. If for coefficients $a_1,a_2,...,a_n$ we have $$a_1x_1+a_2x_2+...+a_nx_n=0$$then $$a_1=a_2=...=a_n=0$$this means that the set of that vectors spans whole the $n-$dimensional space for which any vector in it could be expressed uniquely as a combination of those vectors, meanwhile $$\forall y\qquad,\qquad \exists \{a_i\}\to y=a_1x_1+a_2x_2+...+a_nx_n\\\text{if }y=b_1x_1+b_2x_2+...+b_nx_n\text{ then }a_i=b_i$$

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