 # We have linearly independent vectors e_1,e_2,...,e_(m+1) in RR^n. I would like to prove that among their linear combinations, there is a nonzero vector whose first m coordinates are zero. From independence we can get that m+1 <= n. Maybe we can build some system, but what next? trkalo84 2022-09-15 Answered
We have linearly independent vectors ${e}_{1},{e}_{2},...,{e}_{m+1}\in {\mathbb{R}}^{n}$. I would like to prove that among their linear combinations, there is a nonzero vector whose first m coordinates are zero.
From independence we can get that $m+1\le n$. Maybe we can build some system, but what next?
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A linear combination of ${e}_{1},{e}_{2},...{e}_{m}$ is of the form: ${a}_{1}{e}_{1}+{a}_{2}{e}_{2}+...{a}_{m+1}{e}_{m+1}$. If you write each ei out, you see that the condition where the first m coordinates are zero is equivalent to solving an equation of m+1 variables over m equations, which you can always find a solution for.
Another way of saying this is span$\left\{{e}_{1},{e}_{2}...{e}_{m+1}\right\}$ is of dimension m+1, while the vector space of vectors with the first m coordinates zero has dimension n−m. If their intersection consisted only of the 0 vector, dimension of whole space is n−m+m+1, contradiction.

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