linear combination problem dealing with uniqueness

**lllll** · February 9th 2008, 09:04 PM

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

**kalagota** · February 10th 2008, 05:38 AM

Originally Posted by lllll

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

So, let $s \in span \, S$ such that $s = a_1{v_1}+a_2{v_2}+...+a_n{v_n}$ and $s = b_1{v_1}+b_2{v_2}+...+b_n{v_n}$ . Then $a_1{v_1}+a_2{v_2}+...+a_n{v_n} = b_1{v_1}+b_2{v_2}+...+b_n{v_n}$ . This means that $(a_1-b_1){v_1}+(a_2-b_2){v_2}+...+(a_n-b_n){v_n} = 0$ . From the hypothesis, $a_i - b_i = 0, \, \forall i=1,2,...,n$ . Therefore, $a_i = b_i, \, \forall i=1,2,...,n$

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