1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Positive integer proof

  1. Aug 8, 2009 #1
    positive operator proof

    1. The problem statement, all variables and given/known data
    Prove that if T ∈ L(V) is positive, then so is Tk for every positive
    integer k.

    2. Relevant equations

    3. The attempt at a solution
    Let v=b1v1+...+bnvn. Now since T is positive, T has a positive square root. T=S^2. <S^2v, v>=<S^2v1, v>+...+<S^2vn, v>. Now <S^4v, v>=<S^2v1, v>^2+...+<S^2vn, v>^2 yes? Now since S^2 is positive, S^4 is positive. And since S^4 is>=S^2,
    <S^4v1, v>+....+<S^4vn, v> is >=<S^2v1, v>+...+<S^2vn, v> which is >=0.
    But this doesn't show that its true for S^2k. I remember learning about a technique
    called induction. Should that be used?
    Last edited: Aug 8, 2009
  2. jcsd
  3. Aug 8, 2009 #2
    hmmmm. The one you got there looks like it only deals with powers of 2 even if you did prove it by induction.

    Let's get the definitions sorted first. A positive matrix is a matrix with all its elements >0 right?

    The idea behind induction is that you want to prove a base case. This is just another name for the simplest case there is. For example, you would want to prove that this is true for k=1 in this case. And then you consider the k+1 case, for all k >= 1. If you want to learn more, I suggest that you read some actual proofs by induction.

    It sounds like an induction proof but I think you can also just prove this by contradiction. Assume that there exists some k such that T^k is non-positive.
  4. Aug 8, 2009 #3
    Oh ok. Thanks for the help! Btw no offense, but my book said that for T to be positive, <Tv, v> >=0 so T can be a 0 map. But I like your hint, very nice!

    Now we assume that T is positive when k=1. Now if T is postive when k=1,
    then there must be some k where Tk is negative, for k>1.
    Now <Tv, v> is >=0. And <Tkv, v> is <0. But k is >1.
    And T has elements >=0. And Tk=Tk-1T.
    Therefore <Tkv, v> is >=<Tv, v>.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook