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Linear Algebra: Positive Operators

  1. Jul 1, 2008 #1

    LPB

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    1. The problem statement, all variables and given/known data

    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.


    2. Relevant 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>] .


    3. 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 cancelled 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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  3. Jul 1, 2008 #2

    Dick

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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.
     
  4. Jul 2, 2008 #3

    LPB

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    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!
     
  5. Jul 2, 2008 #4

    Dick

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    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 lets you express <xT,y> as a sum of similar expressions where the two vectors are equal.
     
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