# Unique linear combination problem

Let $V$ be a vector space and $S$ a subset of $V$ with the property that whenever $v_1,v_2,\ldots v_n\in S$ and $a_1v_1+a_2v_2+\ldots a_nv_n=0$, then $a_1=a_2=\ldots=a_n=0$. Prove that every vector in the span of $S$ can be expressed uniquely as a linear combination of the vectors of $S$.

My attempt: Let $v\in Span(S)$.

$\Rightarrow v= a_1v_1+a_2v_2+\ldots a_nv_n$

If $a_1v_1+a_2v_2+\ldots a_nv_n=0$, then $a_1=a_2=\ldots=a_n=0$ and the linear combination is unique.

What about when $a_1=a_2=\ldots=a_n\neq 0$?

I can see that in this case $v$ is the linear combination of some $v_i$'s such that not all $a_i$'s are zero. How do I prove the uniqueness?

• Ok, then I'll assume that $v=a_1v_1+a_2v_2+\ldots+a_nv_n$ and $v=b_1v_1+b_2v_2+\ldots+b_nv_n$. So $a_1v_1+a_2v_2+\ldots+a_nv_n=b_1v_1+b_2v_2+\ldots+b_nv_n$. So $(a_1-b_1)v_1+\ldots+(a_n-b_n)v_n$=0. Which gives $a_i-b_i=0$ and so $a_i=b_i$. Am i correct? And thank you :) – Diya Sep 1 '14 at 8:32