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Thread: linear combination problem dealing with uniqueness

  1. #1
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    linear combination problem dealing with uniqueness

    Let $\displaystyle V$ be a vector space and $\displaystyle S \subset V$ with the property that whenever $\displaystyle v_1,v_2,...,v_n \in S$ and $\displaystyle a_1{v_1}+a_2{v_2}+...+a_n{v_n} = 0$, then $\displaystyle a_1 = a_2 = ...=a_n = 0$. Prove that every vector in the span of $\displaystyle S$ can be uniquely written as a linear combination of vectors in $\displaystyle S$.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by lllll View Post
    Let $\displaystyle V$ be a vector space and $\displaystyle S \subset V$ with the property that whenever $\displaystyle v_1,v_2,...,v_n \in S$ and $\displaystyle a_1{v_1}+a_2{v_2}+...+a_n{v_n} = 0$, then $\displaystyle a_1 = a_2 = ...=a_n = 0$. Prove that every vector in the span of $\displaystyle S$ can be uniquely written as a linear combination of vectors in $\displaystyle S$.
    So, let $\displaystyle s \in span \, S$ such that $\displaystyle s = a_1{v_1}+a_2{v_2}+...+a_n{v_n}$ and $\displaystyle s = b_1{v_1}+b_2{v_2}+...+b_n{v_n}$. Then $\displaystyle a_1{v_1}+a_2{v_2}+...+a_n{v_n} = b_1{v_1}+b_2{v_2}+...+b_n{v_n}$. This means that $\displaystyle (a_1-b_1){v_1}+(a_2-b_2){v_2}+...+(a_n-b_n){v_n} = 0$. From the hypothesis, $\displaystyle a_i - b_i = 0, \, \forall i=1,2,...,n$. Therefore, $\displaystyle a_i = b_i, \, \forall i=1,2,...,n$
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