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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?

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Make it, assuming you have two linear combinations.

(If you want the whole demonstration, ask it in comment, will edit.)

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  • $\begingroup$ Why by contradiction? It's a direct proof. $\endgroup$ – egreg Sep 1 '14 at 8:29
  • $\begingroup$ 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 :) $\endgroup$ – Diya Sep 1 '14 at 8:32
  • $\begingroup$ Yeah, correct. Well done. $\endgroup$ – Deuteu Sep 1 '14 at 8:34
  • $\begingroup$ @egreg In my mind that is by contradiction because you assume that you have two different combinations but you conclued that it's not possible. But indeed, you can assume that you have two linear combinations and conclued that they are the same, that's more logic maybe. :) $\endgroup$ – Deuteu Sep 1 '14 at 8:39
  • $\begingroup$ @Deuteu The attempted proof uses no contradiction, does it? $\endgroup$ – egreg Sep 1 '14 at 8:40

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