trkalo84

2022-09-15

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?

From independence we can get that $m+1\le n$. Maybe we can build some system, but what next?

Simon Zhang

Beginner2022-09-16Added 7 answers

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$\{{e}_{1},{e}_{2}...{e}_{m+1}\}$ 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.

Another way of saying this is span$\{{e}_{1},{e}_{2}...{e}_{m+1}\}$ 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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