# Linear Algebra/ Linearly Independent

## 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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Dick
Homework Helper

## 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?

True, obviously I need to think this through a little more.

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?

Dick
Homework Helper
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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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?

Dick
Homework Helper
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.

OK (this will be a good test in patience for you)

suppose x=(1,1) and T(x)=(-1,-1)

Dick
Homework Helper
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?

Dick
Homework Helper
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.

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!

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?

Dick
Homework Helper
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.

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?

Dick
Homework Helper
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?

apply T^(m-2) to (c1)T^m(x)+... ??

Dick
Homework Helper
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.

apply T^(m-2) to (c1)T^m(x)+... ??

Dick
Homework Helper
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.

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?

Dick
Homework Helper
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?