# Proof of linear dependence

1. Oct 16, 2005

### tandoorichicken

This may be a really simple proof but its giving me grief.

If {$v_1, v_2, v_3$} is a linearly dependent set of vectors in $\mathbb{R}^n$, show that {$v_1, v_2, v_3, v_4$} is also linearly dependent, where $v_4$ is any other vector in $\mathbb{R}^n$.

Any hints on where to start? I started out by writing out all the claims that could be made by taking the first set of vectors to be linearly independent, but that didn't get me terribly far.

2. Oct 16, 2005

### TD

The definition of lineair dependance (or at least an equivalent statement) is that you can form the zero-vector with a lineair combination of your vectors without having all 0 coefficients. Now, you can do that for your first set, would you be able to do it too for the second one?

3. Oct 16, 2005

i don't like this question because it doesn't exlude a v_4 that is in the span of the first set of vectors.

in that case, the second set WOULD be dependent!

:grumpy:

4. Oct 16, 2005

### Tzar

But thats exaclty what you want to prove, so your proof is finished!

5. Oct 16, 2005

### tandoorichicken

I think I got it. The rule basically says the weights c_1, c_2, ... c_n can't all be equal to zero at the same time. If the first set is already linearly dependent, i.e. $c_1\vec{v}_1 + c_2\vec{v}_2 + c_3\vec{v}_3 = 0$, then we can take c_4 = 0, which would make $c_1\vec{v}_1 + c_2\vec{v}_2 + c_3\vec{v}_3 + c_4\vec{v}_4 = 0$ as well, therefore the second set is also linearly dependent.

6. Oct 17, 2005

### TD

That's what I meant

7. Oct 17, 2005

oh!

i thought it said INdependent!

:rofl:

8. Oct 17, 2005

yep, that's it. good job.