Prove that there exists a combination among $m+1$ independent vectors in $\mathbb R^n$ whose first $m$ coordinates are zero.

61 Views Asked by At

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 $m+1 \leq n$. Maybe we can build some system, but what next?

1

There are 1 best solutions below

0
On BEST ANSWER

A linear combination of $e_1, e_2, ... e_m$ is of the form: $a_1e_1 + a_2e_2 + ...a_{m+1}e_{m+1}$. If you write each $e_i$ 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.