# Linear Algebra- Dependent or independent

1. Apr 25, 2016

### MozAngeles

1. The problem statement, all variables and given/known data
Let(v1, v2, v3) be three linearly independent vectors in a vector space V. Is the set {v1-v2, v2-v3, v3-v1} linearly dependent or independent?

2. Relevant equations
Linearly independent is when c1v1+c2v2+...+ckvk=0
and c1=c2=...ck=0
3. The attempt at a solution
c1(v1-v2)+ c2(v2-v3)+ c3(v3-v1)=0

c1-c3=0
-c1+c2=0
-c2+c3=0

therefore c1=c2=c3 and since c1, c2 and c3 are zero because for the first set of independent vectors I got c1v1+c2v2+c3v3=0 all c1=c2=c3=0,
which means this is the case for the second set and it must be linearly independent.

This is what i got but my answer key says the second set is linearly dependent. I'm having trouble seeing why.
Thanks for any help

2. Apr 25, 2016

### Samy_A

Can you explain in detail how you concluded from c1=c2=c3 that c1, c2 and c3 are zero?

3. Apr 25, 2016

### MozAngeles

from my set of equations i concluded that c1=c2=c3. So then I had made the assumption that c1=c2=c3 was zero because from the given set v1, v2, v3 being linearly indepedent, which would mean that c1v1+c2v2+c3v3=0 must have c1=c2=c3 from the definition of linear dependence. I assumed they were the same constants, is this wrong to assume?

4. Apr 25, 2016

### Samy_A

$v_1,v_2,v_3$ being linearly independent means that if $d_1v_1+d_2v_2+d_3v_3=0$, then $d_1=0,d_2=0,d_3=0$.
There is no reason whatsoever to assume that the numbers $c_1,c_2,c_3$ that you chose for the set {$v_1-v_2, v_2-v_3, v_3-v_1$} must also work for $v_1,v_2,v_3$.

5. Apr 25, 2016

### MozAngeles

but even still for my set of {v1-v2,v2-v1....} the c1=c2=c3 are all equall and I had set my equations to equal zero, then the only way this will be true is if they all equal zero, right?

6. Apr 25, 2016

### Samy_A

Why? All you found is that $c_1=c_2=c_3$.
That seems sufficient to have $c_1(v_1-v_2)+c_2(v_2-v_3)+c_3(v_3-v_1)=0$.

7. Apr 25, 2016

### MozAngeles

Ok, I'm seeing it clearer now. Thank you

8. Apr 25, 2016

### Staff: Mentor

I would add that because the vectors are linearly independent, there can be no other solutions for the constants $d_1, d_2,$ and $d_3$.

If the three vectors were linearly dependent, the equation $d_1v_1+d_2v_2+d_3v_3=0$ would have an infinite number of solutions for the constants, including $d_1=0,d_2=0,d_3=0$. This is a fine point that often eludes new students of linear algebra.