# Linear algebra proof

1. May 14, 2016

### Danielm

1. The problem statement, all variables and given/known data
Prove the following theorem: Let (v1, . . . , vk) be a sequence of vectors from a vector space V . Prove that the sequence if linearly dependent if and only if for some j, 1 ≤ j ≤ k, vj is a linear combination of (v1, . . . , vk) − (vj ).

2. Relevant equations

3. The attempt at a solution
the if and only if is what bothers me. I know how to prove the following direction: If vj is a linear combination of (v1,.....,vk)-vj then linearly dependent

My approach is if c1v1+.....+ckvk=vj then c1v1+....+ckvk-vj=0, so there exists a set of constants c1,..,ck,cj=-1 such that c1v1+...+ckvk=0 (note c1,...,ck can't all be zero) is that right?

I don't know how to show if linearly dependent then vj is a linear combination of (v1,.....,vk)-vj, I guess the contrapositive would be ok

if for all vj, vj is not a linear combination of (v1,...,vk)-vj then the sequence of vectors is not linearly independent.

Assume to the contrary that there exists a vj such that vj is a linear combination of (v1,..,vk)-vj and the sequence of vectors is not linearly independent.

then there exists a set of constants such that c1v1+.....+ckvk=vj ,so c1v1+.....+ckvk-vj=0 which shows the system is linearly dependent, so contradiction
..

Last edited: May 14, 2016
2. May 14, 2016

### andrewkirk

If $v_1,...,v_k$ are linearly dependent then there exist $c_1,...,c_k$ such that $\sum_{i=1}^kc_iv_i=0$. You want to be able to choose one of the vectors in that sum to be your $v_j$ and then rearrange the equation so that $v_j$ is all by itself on one side of the equation and does not appear on the other side. What property does $c_j$ have to have to allow you to do that? Can you be sure that at least one of the coefficients has that property? Why?

3. May 14, 2016

### Danielm

vj is a linear combination of (v1,.....,vk)-vj.

c1v1+...+c_j-1v_j-1+c_j+1v_j+1+.....+c_kv_k-v_j=0

so c1v1+...+c_j-1v_j-1+c_j+1v_j+1+.....+c_kv_k has to add up to v_j so the sum is 0 and this can't be achieved if all c_j are zero

4. May 14, 2016

### andrewkirk

Right. And if there is no linear combination that adds to zero for which the $c_j$ are not all zero, what does that tell us about whether the set of vectors is linearly dependent?

5. May 14, 2016

### Danielm

it means the set of vectors is linearly independent hence the only solution to the system is the trivial solution

6. May 14, 2016

### andrewkirk

Good. Do you now understand how to prove the 'only if' direction?

7. May 14, 2016

### Danielm

yes, well what I understand about the bi-conditional is that we have to prove both directions of the statement. If vj is a linear combination of (v1,.....,vk)-vj then linearly dependent. f linearly dependent then vj is a linear combination of (v1,.....,vk)-vj. The second one I would prove it by using the contrapositive. if for all vj, vj is not a linear combination of (v1,...,vk)-vj then the sequence of vectors is not linearly independent. And then I would use proof by contradiction which is basically the same thing as the first direction.