Linearly independence proof

  • #1
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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?
 

Answers and Replies

  • #2
LCKurtz
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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?
 
  • #3
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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?
 
  • #4
LCKurtz
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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 ...?
 
  • #5
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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
 
  • #6
LCKurtz
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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##.
 
  • #7
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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.
 
  • #8
LCKurtz
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Oh I see what you're saying. c3 can be something like 2, then c1=c2=-2. Thus the system is linearly dependent [STRIKE]for this case[/STRIKE].
Right, because [I presume] you calculated ##-2w_1-2w_2+2w_3## and got zero. And you wouldn't say "for this case". They are linearly dependent period and these choices of the constants are one way of proving it.
 

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