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Linear Independence

  1. Apr 16, 2006 #1
    Here's a simple question that I can't seem to get:

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

    I know that [tex]m\leq \dim V[/tex] and [tex]v,Tv,...,T^{m-1}v[/tex] are all nonzero.
     
  2. jcsd
  3. Apr 16, 2006 #2
    What happens if [tex]0 = a_1 v + a_2Tv+ ... + a_mT^{m-1}v[/tex]? You want to show that all a_i are 0 and you have two bits of information:
    (1) T is linear
    (2) [tex]T^mv=0[/tex]
    How can you use this to get information about a_1?
    Also, do you know what T(0) is?
     
  4. Apr 16, 2006 #3
    I was just about to post that I got it. T(0) is not used it my proof. Thanks anyway.
     
  5. Apr 17, 2006 #4

    HallsofIvy

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    If, as I assume, T is a linear transformation then T(0)= 0 and there would be no reason to use it!
     
  6. Apr 17, 2006 #5

    0rthodontist

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    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.
     
  7. Apr 17, 2006 #6

    matt grime

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    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).
     
    Last edited: Apr 17, 2006
  8. Apr 17, 2006 #7

    0rthodontist

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    Well, every time you multiply you have to say on the left T(0) = 0 = (rest of expression).
     
    Last edited: Apr 17, 2006
  9. Apr 17, 2006 #8

    matt grime

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    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?
     
  10. Apr 17, 2006 #9
    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.
     
  11. Apr 17, 2006 #10

    matt grime

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    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.
     
  12. Apr 17, 2006 #11
    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.
     
    Last edited: Apr 17, 2006
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