How can a basis in a vector space be used to determine linear independence?

[Krovski]

From what I understand, a basis is essentially a subset of a vector space over a given field.

Now what I'm not so sure of is the linearly independence part. If the basis has two linearly independent vectors, then than means they aren't collinear: rather, they wouldn't have the same slope or be generated by each other?

let's say a vector v$\epsilon$V is (v$_{1}$, v$_{2}$,...,v$_{n}$)
and a vector w$\epsilon$V is (w$_{1}$, w$_{2}$,...,w$_{n}$)
such that
v $\neq$ cw.
for any constant c$\epsilon$$\textbf{F}$

These would then be non-collinear which means there are no linear operators that can turn v into w?

Any insight would be greatly appreciated and thank you in advance.

 

[tiny-tim]

Hi Krovski!

These would then be non-collinear which means there are no linear operators that can turn v into w?

Yes.

Similarly, 3 vectors u v and w are linearly independent if there are no linear operators that can turn u and v into w, ie au + bv = w (and similarly for any number of vectors).

 

[wisvuze]

Actually, there are no *linear combinations* of u and v that can give you w. But you can still have a linear operator that could possibly take u to v.

For example, suppose
{ u , v , w } is a basis ( for a 3 dimensional space, and they are all linearly independent ) then permuting this basis to { v , u , w } and extending by linearity is a linear operator