# Proof of linear dependence

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.

## Answers and Replies

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TD
Homework Helper
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?

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:

in that case, the second set WOULD be dependent!
:grumpy:
But thats exaclty what you want to prove, so your proof is finished!

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.

TD
Homework Helper
That's what I meant

oh!

i thought it said INdependent!

:rofl:

tandoorichicken said:
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.
yep, that's it. good job.