Proving Linear Independence of (v,Tv,...,T^{m-1}v): A Solution

[Treadstone 71]

Here's a simple question that I can't seem to get:

"Suppose for some v $$T^{m-1}v\neq 0$$ and $$T^mv=0$$. Prove that $$(v,Tv,...,T^{m-1}v)$$ is linearly independent."

I know that $$m\leq \dim V$$ and $$v,Tv,...,T^{m-1}v$$ are all nonzero.

 

[Euclid]

What happens if $$0 = a_1 v + a_2Tv+ ... + a_mT^{m-1}v$$? You want to show that all a_i are 0 and you have two bits of information:
(1) T is linear
(2) $$T^mv=0$$
How can you use this to get information about a_1?
Also, do you know what T(0) is?

 

[Treadstone 71]

I was just about to post that I got it. T(0) is not used it my proof. Thanks anyway.

 

[HallsofIvy]

I was just about to post that I got it. T(0) is not used it my proof. Thanks anyway.

If, as I assume, T is a linear transformation then T(0)= 0 and there would be no reason to use it!

 

[0rthodontist]

Well, there is a proof that depends on T(0) (with T a square matrix). If you left-multiply by T enough times you will get a1T^(m-1)v = 0 showing that a1 = 0. Then from that, if you left-multiply one fewer time you can get a2 = 0, and so on.

 

[matt grime]

How does that utilize T(0), which is 0 since T is a linear map? And T is necessariyl square if we are forced to pick a basis (which we should never do unless we have to; it is a crutch to understanding) since T is in End(V) (as is contextually clear; we cannot apply T to Tv otherwise).

 

[0rthodontist]

Well, every time you multiply you have to say on the left T(0) = 0 = (rest of expression).

 

[matt grime]

Ah, I see what you're saying. But since T is a linear map I wouldn't have even mentioned this explicitly, which is probably what is puzzling me (and HallsOf Ivy): why (did Euclid) even mention this?

 

[Euclid]

I only mentioned it because I wasn't sure if it was a well-known fact to Treadstone. The proof I was thinking of would have involved applying T repeatedly to both sides of that equation above. If you weren't aware that T(0)=0, then you might not have noticed this proof.

 

[matt grime]

Hopefully it is a well known fact to everyone (who knows what a linear map is), since it is part of the defining charactistic of being linear. Of course I've always been too idealistic about things like that.

 

[Euclid]

You'd be surprised. I was a TA for linear algebra, and I was shocked by how many students (at the end of the term) still didn't know basic facts about linear maps.