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Let $V$ be a vector space over a field $K$ and let $A = \{c_{1} , c_{2} , ...., c_{n}\}$ be a basis for $V$. If $m > n$ and $a_{1} , a_{2} , a_{3} ...., a_{m}$ are elements of $V$ , how to prove that these elements are linearly dependent?

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You might want to check out Axler's "Linear Algebra Done Right", pages 25 - 30 for an excellent treatment of the relationship between a basis and a spanning list (or a list longer than the length of the basis) and linear (in)dependence. It's a very good book in general. – TheBirdistheWord Sep 27 '11 at 1:49
Use Steinitz's theorem. – a.r. Sep 27 '11 at 13:43
Dear Prigmin: My first reaction would be to ask you: What exactly do you assume as known? – Pierre-Yves Gaillard Sep 27 '11 at 13:59
up vote 1 down vote accepted

Steitnitz's theorem says: is you have a basis $c_1, \dots , c_n$ of a vector space $V$ and $a_1, \dots , a_m$ are linearly independent vectors, then you can substitute $m$ vectors of the basis $c_1, \dots , c_n$ by $a_1, \dots , a_m$ getting a new basis. In particular, $m\leq n$.

Since you say that $m> n$, then $a_1 , \dots , a_m$ must be linearly dependent.

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Assume that you have a linear combination of the $a_n$'s. By expressing everything up in coordinates with respect to the basis, you get a homogenous system of $n$ linear equations in $m$ unknowns (the unknowns being the coefficients of the linear combination). If this system has a nonzero solution, then the $a$'s are linearly dependent ...

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Thanks to you ... – Prigmin Smith Sep 27 '11 at 19:02

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