# Positive integer proof

1. Aug 8, 2009

### evilpostingmong

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. Aug 8, 2009

### aostraff

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.

3. Aug 8, 2009

### evilpostingmong

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>.