# Thread: Span and linearly independence question

1. ## Span and linearly independence question

Suppose that {u; v} is a linearly independent set of vectors and w is not in the span of {u; v}. Show that the set {u, v,w} is linearly independent.

So we no that a linearly independent set:
x1v1+x2v2+.......xpvp=0 where x has only the trivial soltuion
So does that mean i have to plug in u and w into that equation to show it will equal 0?
Im confused on how to show it when w is not in the span. Any Help would be appriciated

2. Originally Posted by kensington
Suppose that {u; v} is a linearly independent set of vectors and w is not in the span of {u; v}. Show that the set {u, v,w} is linearly independent.

So we no that a linearly independent set:
x1v1+x2v2+.......xpvp=0 where x has only the trivial soltuion
So does that mean i have to plug in u and w into that equation to show it will equal 0?
Im confused on how to show it when w is not in the span. Any Help would be appriciated

Supose $au+bv+cw=0\,,\,\,a,b,c\mbox{ scalars }\Longrightarrow cw = -au-bw$ . If $c\neq 0$ then

dividing by $c$ we'd get that w is a lin. comb. of u,v, so it must be $c=0\Longrightarrow au+bv=0\Longrightarrow \mbox{ also } a=b=0\Longrightarrow$ we're done.

Tonio

3. in a different way,
Let $a,b,c$ be scalars such that $au+bv+cw=0$
if we had $c\neq 0$ we would have $w=-\frac{1}{c}\left ( au+bv \right )$ which implies $w\in \texttt{Span}(v,u)$ and that's not true by assumption.Therefore $c=0$.As a result, we go back to $au+bv=0$.And.. we're done.