Is it possible to prove linear dependence using a simple approach?

[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.

 

[TD]

The definition of lineair dependence (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?

 

[Brad Barker]

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!

 

[Tzar]

in that case, the second set WOULD be dependent!

But that's exactly what you want to prove, so your proof is finished!

 

[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.

 

[TD]

That's what I meant

 

[Brad Barker]

oh!

i thought it said INdependent!

 

[Brad Barker]

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.