Show that a linear map is linearly independent

[Mitch_C]

Homework Statement

Let f:V\rightarrow V be a linear map and let v\inV be such that
f^n(v)\neq0 and f^(n+1)(v)=0. Show that v,f(v),...,f^(n-1)(v) are linearly independent.

The Attempt at a Solution

I'm really stuck with this one. I know the definition of linear independence and I can see why this might be the case but I don't know how to go about showing this. If anyone could point me in the right direction I should be okay.

thanks

 

[JonF]

Can you show that for m<n f^m(v)≠0 and that for p>n f^p(v)=0. I.e. that fⁿ⁺¹ has to be the first mapping that sends v to 0?

If so could you then apply some f^b to both sides of this equation: 0 = α₀v +α₁f(v) + α₂f²(v) + ... + α_nf^n-1(v) that would show α₀ = 0 (hint pick the right b). Could you do something similar for α₁ = 0, etc?

 

[Mitch_C]

Can you show that for m<n f^m(v)≠0 and that for p>n f^p(v)=0. I.e. that fⁿ⁺¹ has to be the first mapping that sends v to 0?

If so could you then apply some f^b to both sides of this equation: 0 = α₀v +α₁f(v) + α₂f²(v) + ... + α_nf^n-1(v) that would show α₀ = 0 (hint pick the right b). Could you do something similar for α₁ = 0, etc?

Ok I think I get you. So how would I go about picking an m and p? Or can I prove that arbitrarily? and for the f^b would that be so f^b(v) = 1?

 

[JonF]

Prove it arbitrarily :

The p is pretty easy. Play around with fⁿ⁺²(v) = f(fⁿ⁺¹(v)). That should show you why p > n goes to zero.

The m<n goes isn't zero is a bit trickier. I don’t want to give away the answer, so let’s consider a nice case where n+1 = 5. So f⁵(v) = 0, f⁴(v) ≠ 0, could f³(v) = 0 if f(f³(v)) = f⁴(v) ≠ 0

Using these properties see what happens if you let b=n and apply it to that equation I gave you.

and for the fb would that be so fb(v) = 1?

No, you need to show that the only way that equation can =0 is if all of the α_n terms are 0. That’s what it means to be linearly independent. With this method we're going to pick them off one at a time. starting with α₀ all the way to α_n-1


EXTRA HINT: fⁿ(0) = 0 for any n

 

[Mitch_C]

Ok that's great thanks a lot! Not only do I now have the question done but I understand it all too! I'm a happy bunny :)

So to make sure I did it right, I took fⁿ the first time to show a₀=0 and then f^n-1 to show a₁=0 and so on until you get f^n-n which just gives you the equation back. And because all of the other a_is are zero and f^n-1(v) is different from zero then the final one a_n=0. So they're linearly independent and they all lived happily ever after!

Thanks a lot!

 

[JonF]

exactly! did you get the p and m part? that's a key step.

 

[Mitch_C]

Yeah I got that part too. I also see why it's a key part :)