Linear Algebra/ Linearly Independent

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Homework Statement



Let T be a linear transformation of a vector space V into itself. Suppose x ε V is such that Tm(x)=0, Tm-1(x) not equal 0 for some positive integer m. show that x, T(x), …, Tm-1(x) are linearly independent.

In regards to Tm and Tm-1 m and m-1 are upperscript on the T. I don't know how to make it do that on this site.




The Attempt at a Solution



To show linearly independent, show that the only linear combination of transformations that = 0 is the one in which the coefficients are zero. For every xεV and cεR, real numbers, Suppose that c1, c2, ...,cm are scalars s.t. (c1)x + (c2)T(x) + ...(cm)Tm-1(x)=0 then
Σci Tm-1(x) =0 i=1 to m which equals Σci ΣTm-1(x) =0 Therefore all c's =0 hence linearly independent. Tm(x) could not be included since it equals 0 which means c could be nonzero.

This doesn't seem right (too easy)???
 

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  • #2
Dick
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Homework Statement



Let T be a linear transformation of a vector space V into itself. Suppose x ε V is such that Tm(x)=0, Tm-1(x) not equal 0 for some positive integer m. show that x, T(x), …, Tm-1(x) are linearly independent.

In regards to Tm and Tm-1 m and m-1 are upperscript on the T. I don't know how to make it do that on this site.




The Attempt at a Solution



To show linearly independent, show that the only linear combination of transformations that = 0 is the one in which the coefficients are zero. For every xεV and cεR, real numbers, Suppose that c1, c2, ...,cm are scalars s.t. (c1)x + (c2)T(x) + ...(cm)Tm-1(x)=0 then
Σci Tm-1(x) =0 i=1 to m which equals Σci ΣTm-1(x) =0 Therefore all c's =0 hence linearly independent. Tm(x) could not be included since it equals 0 which means c could be nonzero.

This doesn't seem right (too easy)???
I don't see anything in there that looks even vaguely correct. Take the case m=2. Then you know T^2(x)=0 and you want to look at c1*x+c2*T(x)=0. Suppose you operate on that equation with T?
 
  • #3
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True, obviously I need to think this through a little more.
 
  • #4
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What if I get rid of the Σ symbols and say c1*x + c2*T(x) + .... cm* T^m-l(x) but before I go further, since you said you didn't see anything vaguely correct, I'm afraid I'm not even approaching this from the right direction?
 
  • #5
Dick
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What if I get rid of the Σ symbols and say c1*x + c2*T(x) + .... cm* T^m-l(x) but before I go further, since you said you didn't see anything vaguely correct, I'm afraid I'm not even approaching this from the right direction?
Sure, it's probably easier to read c1*x+...+cm*T^(m-1)(x) than the sigma notation. And no, I don't think your direction is correct. Like I said in my last post. Try thinking about the case m=2 first. It should make things a lot clearer.
 
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  • #6
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ok so nothing seems clear to me :( I did read in a definition that any subset of a vector space that contains the zero vector is linearly dependent. So that would be the reason the T^m(x) could not be included.

for the case m=2 where c1*x + c2*T(x) = 0 I'm afraid to ask this but here goes,

what if you choose x=1 and then T(1) could be -1 (since T^1 not equal 0 it could be negative?) then c1 +(-c2) =0 then c1=c2 which is not the trivial solution 0.

From my original question, I know this is wrong, but why?
 
  • #7
Dick
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ok so nothing seems clear to me :( I did read in a definition that any subset of a vector space that contains the zero vector is linearly dependent. So that would be the reason the T^m(x) could not be included.

for the case m=2 where c1*x + c2*T(x) = 0 I'm afraid to ask this but here goes,

what if you choose x=1 and then T(1) could be -1 (since T^1 not equal 0 it could be negative?) then c1 +(-c2) =0 then c1=c2 which is not the trivial solution 0.

From my original question, I know this is wrong, but why?
It's wrong because x and T(x) are vectors. Not numbers. If they were just numbers, then you would be right and they couldn't possibly be independent.
 
  • #8
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OK (this will be a good test in patience for you)

suppose x=(1,1) and T(x)=(-1,-1)
 
  • #9
Dick
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OK (this will be a good test in patience for you)

suppose x=(1,1) and T(x)=(-1,-1)
They aren't linearly independent, are they?
 
  • #10
Dick
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OK (this will be a good test in patience for you)

suppose x=(1,1) and T(x)=(-1,-1)
You are asking the right kind of questions, because they are showing you really don't understand what linear independence is. And that's a good thing, believe it or not. You have a good example of two vectors that are NOT linearly independent. Now try and show that (1,1) and (-1,1) ARE linearly independent. This is a ways from your original question, but once you figure out what linear independence actually means, it will be a lot easier.
 
  • #11
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Thank you for being patient with me. In your first reply you mentioned to operate on that equation with T, which is what my professor suggested to our class. So, I am going to spend a little time in thought on that and try it again.

I think my confusion with linear independence is when it's with transformations? Hopefully a little more study will help.
Thanks again!
 
  • #12
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OK, If you apply T to (c0)x + (c1)T(x) + ...(cm-1)T^m-1(x)=0 you would get
(c0)T(x) + (c1)T^2(x) + ...+ (cm-2)T^m-1(x) +(cm-1)T^m(x)=T(0)=0 then I'm stuck again. Can you help me from here?
 
  • #13
Dick
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OK, If you apply T to (c0)x + (c1)T(x) + ...(cm-1)T^m-1(x)=0 you would get
(c0)T(x) + (c1)T^2(x) + ...+ (cm-2)T^m-1(x) +(cm-1)T^m(x)=T(0)=0 then I'm stuck again. Can you help me from here?
In this case you are better off applying T^(m-1) to that expression.
 
  • #14
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ok then (c0)T^m-1(x) +(c1)T^m(x) +...+ (ck)T^m-1+k +... + (cm-1)T^m-1+m-1 =0

so (c0)T^m-1(x)=0 and c0 = 0 by Induction Principle (multiplying by T^m-k-1) all ck =0 and the set is linearly independent.


My classmate found this proof in a book. He didn't understand it and I don't either.

what happened to the (c1)T^m(x) the c1 wouldn't have to be 0?
 
  • #15
Dick
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ok then (c0)T^m-1(x) +(c1)T^m(x) +...+ (ck)T^m-1+k +... + (cm-1)T^m-1+m-1 =0

so (c0)T^m-1(x)=0 and c0 = 0 by Induction Principle (multiplying by T^m-k-1) all ck =0 and the set is linearly independent.


My classmate found this proof in a book. He didn't understand it and I don't either.

what happened to the (c1)T^m(x) the c1 wouldn't have to be 0?
When you apply T^(m-1) you get c0*T^(m-1)(x)=0 since all of the higher powers of T applied to x are zero. Since T^(m-1)(x) is nonzero, that means c0=0. So substitute c0=0 and then apply T^(m-2). What do you conclude about c1?
 
  • #16
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apply T^(m-2) to (c1)T^m(x)+... ??
 
  • #17
Dick
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apply T^(m-2) to (c1)T^m(x)+... ??
No. Apply it to (c1)T(x)+...=0. The original expression with c0 set to 0.
 
  • #18
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apply T^(m-2) to (c1)T^m(x)+... ??
 
  • #19
Dick
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apply T^(m-2) to (c1)T^m(x)+... ??
No. Apply it to (c1)T(x)+...=0. The original expression with c0 set to 0.
 
  • #20
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then c1=0?? since applying T^(m-2) to (c1)T^(x) gives you (c1)T^(m-1)x

I'm not sure how to ask this or if you can even explain in writing but why do we keep applying T?
 
  • #21
Dick
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then c1=0?? since applying T^(m-2) to (c1)T^(x) gives you (c1)T^(m-1)x

I'm not sure how to ask this or if you can even explain in writing but why do we keep applying T?
You start with:
(c0)x+(c1)T(x) + (c2)T^2(x) + ...+ (cm-1)T^m-1(x)=0
You want to show those vectors are linearly independent. That means you want to show all of the c's must be 0. If you operate with T^(m-1) that gets rid of everything except for c0*T^(m-1)(x)=0. So you conclude c0=0. Put that into the first expression. So now you have:
(c1)T(x) + (c2)T^2(x) + ...+ (cm-1)T^m-1(x)=0
Operate with T^(m-2). That kills everything except for c1*T^(m-1)=0. So you conclude c1=0. Etc Etc.
 
  • #22
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OK, I think my light bulb came on! You are an excellent teacher! You deserve another award under your name.

Thank you so much for all of your patience!
 

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