# Linearly Independent Vectors

• Nov 8th 2008, 06:01 PM
horacejerry
Linearly Independent Vectors
Let a system of vectors $\displaystyle \bold{v}_{1},\bold{v}_{2}, \ldots, \bold{v}_{r}$ be linearly independent but not generating. Show that it is possible to find a vector $\displaystyle \bold{v}_{r+1}$ such that the system $\displaystyle \bold{v}_{1}, \bold{v}_{2}, \ldots, \bold{v}_{r}, \bold{v}_{r+1}$ is linearly independent.

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

To show linear independence, we have to show that $\displaystyle \sum_{i=1}^{k+1} \alpha_{k} \bold{v}_{k} = \bold{0}$ with $\displaystyle \alpha_{k} = 0$.
• Nov 9th 2008, 04:14 AM
PaulRS
$\displaystyle \sum\limits_{k = 1}^{n + 1} {\alpha _k \cdot \bold{v_k} } = \bold{0}$

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