# Thread: Linearly Independent Vectors

1. ## Linearly Independent Vectors

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 $\alpha_{k} = 0$.

2. $
\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!