Linearly Independent Vectors

**horacejerry** · November 8th 2008, 06:01 PM

Let a system of vectors $\bold{v}_{1},\bold{v}_{2}, \ldots, \bold{v}_{r}$ be linearly independent but not generating. Show that it is possible to find a vector $\bold{v}_{r+1}$ such that the system $\bold{v}_{1}, \bold{v}_{2}, \ldots, \bold{v}_{r}, \bold{v}_{r+1}$ is linearly independent.

So let $\bold{v}_{r+1}$ be any vector that cannot be represented as a linear combination $\sum_{k=1}^{r} \alpha_{k} \bold{v}_{k}$ . So $\bold{v}_{1}, \bold{v}_{2}, \ldots, \bold{v}_{r}, \bold{v}_{r+1}$ does not form a basis for some vector space $V$ .

To show linear independence, we have to show that $\sum_{i=1}^{k+1} \alpha_{k} \bold{v}_{k} = \bold{0}$ with $\alpha_{k} = 0$ .

**PaulRS** · November 9th 2008, 04:14 AM

$ \sum\limits_{k = 1}^{n + 1} {\alpha _k \cdot \bold{v_k} } = \bold{0} $

By absurd, assume that there is such a linear combination with $ \sum {\left| {\alpha _k } \right|} > 0 $ (1) (that is, not all the scalars are 0). We have that $ \alpha _{n + 1} = 0 $ is impossible by the hypothesis ( the others are LI and we have (1) ), so $ \alpha _{n + 1} \ne 0 $ , and therefore $ \bold{v_{n + 1}} = \sum\limits_{k = 1}^n {\left( -{\tfrac{{\alpha _k }} {{\alpha _{n + 1} }}} \right) \cdot \bold{v_k} } $ that is $ \left[ {\bold{v_1} ,...,\bold{v_n} } \right] $ spans $\bold{v_{n+1}}$ ABSURD!

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