Linear Algebra: Positive Operators

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Homework Help Overview

The discussion revolves around proving that two positive self-adjoint matrices A and B are equal if they satisfy the condition = for all vectors x in a complex vector space. The context is linear algebra, specifically focusing on properties of positive operators in Hilbert spaces.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore whether the equality of the inner products implies the equality of the operators, with some questioning the necessity of the positive condition. Counterexamples are discussed to illustrate potential pitfalls in the reasoning.

Discussion Status

The discussion is active, with participants raising questions about the implications of the given conditions and exploring counterexamples. Some guidance has been offered regarding the nature of self-adjoint operators and the significance of positivity, but no consensus has been reached on the proof approach.

Contextual Notes

Participants note that the operators must be positive and self-adjoint, and there is an exploration of the implications of these properties. The relevance of Lemma 2.12 is acknowledged, but its application remains under discussion.

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Homework Statement



Let A and B be nxn positive self-adjoint matrices such that for all x [tex]\in[/tex] Cn, x*Ax = x*Bx. Prove that A = B. Equivalently, prove that if A, B are positive operators on H such that <Ax,x> = <Bx,x> [tex]\forall[/tex] x [tex]\in[/tex] H, then A = B. Hint: See Lemma 2.12.


Homework Equations



Lemma 2.12:
Let H be a Hilbert space over R or C and let T:H [tex]\rightarrow[/tex] H be a self-adjoint linear operator. Then we have:
(i) <Tx,x> is real for all x [tex]\in[/tex] H.
(ii) If H is over R, then for all x,y [tex]\in[/tex] H we have
<Tx,y> = 1/4 [<T(x+y),x+y> - <T(x-y),x-y>].
(iii) If H is over C, then for all x,y [tex]\in[/tex] H we have
<Tx,y> = 1/4 [<T(x+y),x+y> - <T(x-y),x-y>] + i/4 [<T(x+iy),x+iy> - <T(x-iy),x-iy>] .


The Attempt at a Solution



If we already know that <Ax,x> = <Bx,x>, doesn't it automatically follow that A = B? I'm not sure what has to be proven. I went through part (iii) of Lemma 2.12 and plugged in <Ax,x> for <Tx,y>, and everything canceled out so that <Ax,x> = <Ax,x>, but I don't think that this helped me at all. I'm just not sure what I'm being asked to prove...
 
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No, it doesn't follow that <x,Ax>=<x,Bx> implies that A=B. Take the real case. If x=[a,b], A=[[0,1],[0,0]] and B=[[0,0],[1,0]] (I hope the matrix notation is clear), then <x,Ax>=a*b and <x,Bx>=a*b. They are equal. But A is not equal to B. But then the operators A and B are not self adjoint either. Can you come up with an counterexample for the complex case? You can learn a lot by trying to find a counterexample to a theorem if you relax one of the premises.
 
I see what you're saying. Could a simple counterexample for the complex case be A=[[0,i],[0,0]] and B=[[0,0],[i,0]] (keeping x=[a,b])?

If the operator matrices are self-adjoint, it seems A would always equal B ... for instance, A=B=[[0,1],[1,0]] or A=B=[[0,i],[-i,0]]. Is this right? OK, so I see that it makes sense to prove what the question is asking. Only ... how would I start the proof?

And from the question, it seems that A and B must also be positive ... I know that if an operator T is positive, then not only is it self-adjoint, but also <Tx,x> >= 0 ... I don't see how this is essential. Can you give an example of a self-adjoint operator that is not positive? Maybe I'll be able to see the connection then.

Thank you!
 
An example of a matrix that's self-adjoint but not positive is [[-1,0],[0,-1]]. I.e. -I. That's easy, isn't it? I still don't really see what 'positive' has to do with this. Your lemma only requires the matrix to be self-adjoint. And it looks to me like the lemma is all you need. It let's you express <xT,y> as a sum of similar expressions where the two vectors are equal.
 

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