# Linearly independence proof

charlies1902

## Homework Statement

Suppose that S = {v1, v2, v3} is linearly
independent and
w1 = v2
w2 = v1 + v3
and
w3 = v1 + v2 + v3
Determine whether the set T = {w1,w2,w3} is
linearly independent or linearly dependent.

## Homework Equations

Let c1, c2, c3=scalars

c1w1+c2w2+c3w3=0
c1v2+c2v1+c2v3+c3v1+c3v2+c3v3=0
(c2+c3)v1+(c1+c3)v2+(c2+c3)v3=0

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

solving 1st equation gives: c2=-c3
Plug into 3rd gives: -c3+c3=0 → 0=0 what does this mean?

Homework Helper
Gold Member
It means c3 can be anything as long as c1 and c2 are = -c3. in particular not all the c's have to be 0. What does that tell you?

charlies1902
It means T is linearly dependent because in order for it to be independent al c's have to be 0.

So if it had been -c3-c3=0.
c3=-c3
Thus c3=0
That would make it linearly independent right?

Homework Helper
Gold Member
It means T is linearly dependent because in order for it to be independent al c's have to be 0.

So if it had been -c3-c3=0.
c3=-c3
Thus c3=0
That would make it linearly independent right?

Yes. But in this case you can easily find particular values not all zero that work. For example ...?

charlies1902
Yes. But in this case you can easily find particular values not all zero that work. For example ...?

I'm not sure what you are asking. if -c3-c3=0 then c3=0
c2=0 and c1=0

Homework Helper
Gold Member
I'm not sure what you are asking. if -c3-c3=0 then c3=0
c2=0 and c1=0

Woops, misunderstanding of what I meant. I didn't mean you weren't correct. I was referring to your actual problem, where you can find c's not all zero. To finish that problem you should really display three c's that work by plugging them in ##c_1w_1+c_2w_2 + c_3w_3## and getting ##0##.

charlies1902
Oh I see what you're saying. c3 can be something like 2, then c1=c2=-2. Thus the system is linearly dependent for this case.