#### radou

Homework Helper

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I need to check the proof of the proposition below we got for homework, thanks in advance!

Proposition. Let V be a vector space over a field F, and [tex]S = \left\{a_{1}, \cdots, a_{k}\right\}\subset V, k\geq 2[/tex]. If the set S is linearly dependent, and [tex]a_{1} \neq 0[/tex], and if we assume there is an order on S, then there exists at least one element of S which can be shown as a linear combination of its predecessors.

Proof [?]. If S is dependent, then there exists an element in S that can be shown as a linear combination of the rest of vectors from S, so:

[tex]a_{j+1}=\alpha_{1}a_{1}+\cdots+\alpha_{j}a_{j}+\alpha_{j+2}a_{j+2}+\cdots+\alpha_{k}a_{k}[/tex]. (*) Further on, let's assume that the vector [tex]a_{j}[/tex] can be shown as:

[tex]a_{j}=\beta_{1}a_{1}+\cdots+\beta_{j-1}a_{j-1}[/tex]. So, after plugging [tex]a_{j}[/tex] into the equation (*), we get:

[tex]a_{j+1}= \alpha_{1}a_{1}+\cdots+\alpha_{j}(\beta_{1}a_{1}+\cdots+\beta_{j-1}a_{j-1})+\alpha_{j+2}a_{j+2}+\cdots+\alpha_{k}a_{k}[/tex], which implies [tex]a_{j+1}=\gamma_{1}a_{1}+\cdots+a_{j-1}\gamma_{j-1}+\alpha_{j+2}a_{j+2}+\cdots+\alpha_{k}a_{k}[/tex]. We assumed that [tex]a_{j+1}[/tex] is a the combination of all vectors in S, so, since the linear combination does not explicitly contain the vector [tex]a_{j}[/tex], we conclude that [tex]a_{j}[/tex] must be a linear combination of the vectors [tex]\left\{a_{1}, \cdots, a_{j-1}\right\}[/tex] with the coefficients [tex]\gamma_{i}[/tex].

Gee, I have the feeling I missed something big here. :uhh:

P.S. The thread should be called 'Linear dependence proof' or sth like that, but nevermind.

Proposition. Let V be a vector space over a field F, and [tex]S = \left\{a_{1}, \cdots, a_{k}\right\}\subset V, k\geq 2[/tex]. If the set S is linearly dependent, and [tex]a_{1} \neq 0[/tex], and if we assume there is an order on S, then there exists at least one element of S which can be shown as a linear combination of its predecessors.

Proof [?]. If S is dependent, then there exists an element in S that can be shown as a linear combination of the rest of vectors from S, so:

[tex]a_{j+1}=\alpha_{1}a_{1}+\cdots+\alpha_{j}a_{j}+\alpha_{j+2}a_{j+2}+\cdots+\alpha_{k}a_{k}[/tex]. (*) Further on, let's assume that the vector [tex]a_{j}[/tex] can be shown as:

[tex]a_{j}=\beta_{1}a_{1}+\cdots+\beta_{j-1}a_{j-1}[/tex]. So, after plugging [tex]a_{j}[/tex] into the equation (*), we get:

[tex]a_{j+1}= \alpha_{1}a_{1}+\cdots+\alpha_{j}(\beta_{1}a_{1}+\cdots+\beta_{j-1}a_{j-1})+\alpha_{j+2}a_{j+2}+\cdots+\alpha_{k}a_{k}[/tex], which implies [tex]a_{j+1}=\gamma_{1}a_{1}+\cdots+a_{j-1}\gamma_{j-1}+\alpha_{j+2}a_{j+2}+\cdots+\alpha_{k}a_{k}[/tex]. We assumed that [tex]a_{j+1}[/tex] is a the combination of all vectors in S, so, since the linear combination does not explicitly contain the vector [tex]a_{j}[/tex], we conclude that [tex]a_{j}[/tex] must be a linear combination of the vectors [tex]\left\{a_{1}, \cdots, a_{j-1}\right\}[/tex] with the coefficients [tex]\gamma_{i}[/tex].

Gee, I have the feeling I missed something big here. :uhh:

P.S. The thread should be called 'Linear dependence proof' or sth like that, but nevermind.

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