1. ## Proving Linear Dependency

Show that if {v1, v2, v3} is a linearly dependent set of vectors in a vector space V, and v4 is any vector, then {v1, v2, v3, v4} is also linearly dependent.

If I set up a matrix I would end up with a row of 0's in echelon form meaning there is no unique solution, so if I add a 4th column to that matrix would I obtain a row of 0's again, still proving that there is no unique solution, which means it is linearly dependent? If so could somebody set it up for me to see, and if not, could somebody explain how I would go about doing this.

Thanks

2. If $\left\{ {v_1 ,v_2 ,v_3 } \right\}$ is a linearly dependent set in V then $
\alpha _1 v_1 + \alpha _2 v_2 + \alpha _3 v_3 = 0\quad \& \quad \exists \alpha _i \ne 0$
.
Do you understand that?

Well then is it true that $\alpha _1 v_1 + \alpha _2 v_2 + \alpha _3 v_3 + 0v_4 = 0\quad \& \quad \exists \alpha _i \ne 0$.
Does that mean that $\left\{ {v_1 ,v_2 ,v_3, v_4 } \right\}$ is a linearly dependent set in V?

3. So for the first part you're saying a, or alpha in this case, is a scalar that is a real number other than 0, which makes sense, but wouldn't that mean that the set {v1,v2,v3} is linearly independent if they have a unique solution equaling 0 that doesn't involve each of their scalars to be 0?

4. The question is: Do you understand the definition of linear dependent?
What does it mean: $\left\{ {v_1 ,v_2 ,v_3 } \right\}$ is a linearly dependent set in V?

5. I'm not too sure, but I think it means that the vectors can add up to = 0 with the help of scalars.

6. Originally Posted by Keegan
I'm not too sure, but I think it means that the vectors can add up to = 0 with the help of scalars.
what do you mean by "with the help of scalars"? {v_1, v_1, v_3} is linear dependent in V means that there exists some none-zero a_i s.t the sum of a_i*v_i= 0 where i runs from i to 3. What Plato did was he/she fixes a_4=0 so that a_4*v_4=0 then you will have the result from there.