# Linear Independence

1. Feb 24, 2009

### sana2476

I attempted the proof but I don't know how to complete it..

Let u,v,w be linearly independent vectors and x is in <u,v,w>. Then there are unique a,b,y such that x=au+bv+yw

2. Feb 24, 2009

### Tom Mattson

Staff Emeritus
Great, let's see what you've done.

3. Feb 24, 2009

### sana2476

I'm having trouble starting it...if u could help me start it..then i can try to carry it from there

4. Feb 24, 2009

### yyat

Look at the definition of <u,v,w>. What does it mean for x to be in <u,v,w>, spelled out in terms of the definition?
For the uniqueness part, start by assuming that you can write x=au+bv+cw and x=du+ev+fw, then prove that a=d, b=e, c=f.

5. Feb 24, 2009

### sana2476

ok...so if i start the proof by saying if x is in <u,v,w> then there exists d,e,f such that x=du+ev+fw and then if i take the difference, say: (a-d)u+(b-e)v+(c-f)w...would that be right approach?

6. Feb 24, 2009

### Staff: Mentor

By definition, if x is in Span(u, v, w), then there are scalars a, b, and c such that x = au + bv + cw. (I changed letters on you, here.

You want to show that this representation is unique, so one way to do this is to assume the contrary--that the representation is not unique, meaning that there is at least one other way to represent x, say, as du + ev + fw.

Work with these two representations, and you should get a contradiction, which means that your assumption that the representation was not unique must have been incorrect, which gets you back to the representation being unique.