Linear algebra proof

[Danielm]

Homework Statement

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 ).

Homework Equations

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.

Proof by contradiction

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
..

 

[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?

 

[Danielm]

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?

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

 

[andrewkirk]

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

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?

 

[Danielm]

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?

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

 

[andrewkirk]

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

 

[Danielm]

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

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.