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

trkalo84

Answered question

2022-09-15

We have linearly independent vectors e 1 , e 2 , . . . , e m + 1 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 n. Maybe we can build some system, but what next?

Answer & Explanation

Simon Zhang

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.

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