Positive operators

1. Aug 5, 2009

evilpostingmong

1. The problem statement, all variables and given/known data
Prove that the sum of any two positive operators on V is positive.

2. Relevant equations

3. The attempt at a solution
This problem seems pretty simple. But I could be wrong. Should I name two
positive operators T and X such that T=SS* and X=AA*? I have a bad
history of seeing a proof and thinking "wow, this'll be cake" which leads to lazy thinking.
Should I just choose two vectors v1 and v2 and establish two inner products, <SS*v1, v2>+<AA*v1, v2>?
I know that A* is real if X is positive, same goes for S. So <AA*v1, v2>=<Av1, Av2>=<A^2v1, v2>. If I'm
right here, its easy street.

2. Aug 5, 2009

Дьявол

And what is V? Is it vector space ?

3. Aug 5, 2009

evilpostingmong

yes its a vector space. So V is invariant under T (this chapter deals with self-adjoint operators, so I think that this is what is to be assumed).
I have a proof in mind but won't post it if what I said (in the first post ) is wrong.

Last edited: Aug 5, 2009
4. Aug 6, 2009

evilpostingmong

Ok let V be invariant under T. Now let v be an eigenvector under T.
Let V be invariant under X. let v be an eigenvector under X.
Now since T and X are positive, we assume that <Tv, v> >=0 and <Xv, v> >=0 and that
T and X are invariant. Now if <Tv, v> >=0 and <Xv, v>=0, then <(T+X)v, v>=<Tv, v> which
is >=0. If T is a 0 map, <(T+X)v, v>=<Xv, v> which is also >=0. If T and X are not 0 maps, and T and X are both positive, then <(T+X)v, v> is > both <Tv, v> and <Xv, v> since both <Tv, v> and <Xv, v> are >0 if neither are 0 maps. Now if both T and X are 0 maps, then <(T+X)v, v>=0, also >=0.

Now we must show that T+X is self adjoint Since we assume that T and X are positive, T=T* and X=X*. So (T+X)v=Tv+Xv and (T*+X*)v=T*v+X*v. Since Tv=T*v and Xv=X*v, Tv+Xv= T*v+X*v. Therefore T+X is self adjoint.

5. Aug 6, 2009

kof9595995

Ignore me, just here to mark the post and will do it later.

6. Aug 7, 2009

kof9595995

I assume a positive operator is a positive-definite matrix(correct me if I am wrong), do you know what the definition is?

7. Aug 7, 2009

evilpostingmong

You are correct. My book says that a positive definite operator is a positive operator
but didn't use a matrix when describing one. But I assume you're correct in that
by the def of a positive operator: 1. Must be self adjoint 2. <Tv, v> >=0
I would imagine the same to be true for the matrix: its conjugate transpose must equal
the original matrix when dealing with a complex space and its transpose must equal
its original form when dealing with a real space. So its self adjoint. And <Tv, v> >=0.
Which implies all eigenvalues of T must be >=0. Oh and when I said original matrix
I meant the matrix you want to "transpose".

8. Aug 7, 2009

kof9595995

In the 4th post you take v as an eigenvector, I really don't see why,since the definition requires <Tv, v> >0(I think the >=0 case is semi positive definite) for any v, another fallacy in your proof is X and S may not share the same eigenvector. And positive-definite matrix can not be 0 matrix, you don't need to discuss it.
Just let v be any vector, the proof will be OK.

9. Aug 7, 2009

evilpostingmong

Whoops, my bad. As long as the scalar <Tv, v> (say Tv=a1v1+...+anvn where Tv=/=cv where c is some scalar) is >=0, T is positive.
So if I were to go back and revise, just replace "eigenvector" with any vector and the proof
is correct, right? And yes, T and S may not share the same eigenvector.
For example, if T were a 3x3 matrix where every entry on this matrix differ, and
S were a 3x3 identity matrix, its clear that an eigenvector of S is (1 1 0) but
(1 1 0) may not be an eigenvector of T since T(1 1 0) would amount to
(a+b, d+e, 0) where a+b=/=d+e so (a+b, d+e, 0) is not a scalar multiple of (1 1 0).
But we are only focusing on the scalars <Tv, v> and <Sv, v>. So v is arbitrary.
btw this assumes that the basis for V is {(1 1 0), (0 0 1), (0 1 1)}

Last edited: Aug 7, 2009
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