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Difficulty with EASY equivalence statement

  1. Sep 29, 2005 #1
    I want to be able to prove that

    [tex]S=T \, \Leftrightarrow \, \langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle[/tex]

    given that [itex]S,T \in \mathcal{B(H)}[/itex] and [itex]x \in \mathcal{H}[/itex].

    If [itex]S[/itex] and [itex]T[/itex] are bounded linear operators and I suppose that [itex]S = T[/itex] then I have to be able to prove that

    [tex]\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle[/tex]

    Now since [itex]S = T[/itex] then [itex]\|S\| = \|T\|[/itex] and in particular

    [tex]\|Sx\| = \|Tx\| \quad \forall \, x \in \mathcal{H}[/tex].

    Im not really sure how to progress from here. I hope Im not meant to use the Polarization Identity or something like that. Can anyone help?
     
  2. jcsd
  3. Sep 29, 2005 #2

    George Jones

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    Maybe I'm missing something deep, but doesn't this follow trivially from the definition of equality of operators?

    If [itex]S[/itex] and [itex]T[/itex] are bounded linear operators, then

    [tex]S=T \, \Leftrightarrow \, Sx = Tx[/tex]

    for every [itex]x[/itex] in [itex]\mathcal{H}[/itex].

    Regards,
    George
     
  4. Sep 29, 2005 #3
    Right, but surely I can't just say

    Since [itex]S=T \Leftrightarrow Sx=Tx[/itex] then [itex]\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle [/itex].

    This seems too easy. And how would I prove the other way, ie given that [itex]\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle [/itex] prove [itex]S=T[/itex]
     
    Last edited: Sep 29, 2005
  5. Sep 29, 2005 #4

    George Jones

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    Yes.

    Oops, I didn't see the arrow in the other direction - I'll have to think about this. Clearly, [itex]\langle Sx\,|\,y \rangle = \langle Tx\,|\,y \rangle [/itex] for every [itex]x[/itex] and [itex]y[/itex] iff [itex]S=T[/itex], but I don't know if this helps.

    Regards,
    George
     
  6. Sep 29, 2005 #5

    George Jones

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    This doesn't appear to be true, i.e., I think I've found a counterexample.

    Take [itex]\mathcal{H}=\mathbb{R}^2[/itex] with the standard inner product. Let

    [tex]
    S = \left(
    \begin{array}{cc}
    0 & 1\\
    0 & 0
    \end{array}
    \right)
    [/tex]

    and

    [tex]
    T = \left(
    \begin{array}{cc}
    0 & 0\\
    1 & 0
    \end{array}
    \right).
    [/tex]

    Then, [itex]Sx = \left(x_2 , 0)^t[/itex], [itex]Tx = \left(0 , x_1)^t[/itex], and [itex]\langle Sx , x \rangle = \langle Tx , x \rangle = x_{1} x_{2}[/itex] for every [itex]x[/itex].

    Regards,
    George
     
  7. Sep 29, 2005 #6
    Well...as long as you have a Hilbert space (whose ground field is complex) then if two bounded linear operators [itex]S,T \in \mathcal{B(H)}[/itex] then [itex]S = T[/itex] is equivalent to saying that [itex]\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle[/itex]

    This is the exact wording of my problem.
     
  8. Sep 29, 2005 #7

    George Jones

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    It seems that the fact the field is [itex]\mathbb{C}[/itex] is important. It looks like my counterexample doesn't work with [itex]\mathbb{R}^2[/itex] replaced by [itex]\mathbb{C}^2[/itex].

    Regards,
    George
     
  9. Sep 29, 2005 #8
    Indeed George, the field must be complex.

    Anyway, this was all leading to something. I want to actually prove that

    [tex]\langle Sx\,|\,x \rangle = \langle Tx\,|\,x \rangle \quad \Rightarrow \quad \langle Sx\,|\,y \rangle = \langle Tx\,|\,y \rangle[/tex]

    for all [itex]x,y \, \in \mathcal{H}[/itex] and for all [itex]S,T \, \in \mathcal{B(H)}[/itex] where [itex]\mathcal{H}[/itex] is a Hilbert space over [itex]\mathbb{C}[/itex].

    I am afraid that this is much more difficult and I have an idea that it will involve using the Polarization identity. But I need to sleep first.
     
    Last edited: Sep 29, 2005
  10. Sep 29, 2005 #9

    George Jones

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    See this link, page 11 (pdf page 13).

    Regards,
    George
     
  11. Sep 30, 2005 #10
    Thanks for the link George, I like the introduction in that pdf.

    I want to go on what you said about "clearly [itex]\langle Sx\,|\,y \rangle = \langle Tx\,|\,y \rangle[/itex] implies [itex]S=T[/itex]".

    I mean, this seems soooo obvious. But I want to justify it. Is the proof of this implication trivial or not?
     
  12. Sep 30, 2005 #11
    I guess it is not trivial.

    George, your pdf claims that proving this implication requires an EASY application of the Polarization identity.
     
  13. Sep 30, 2005 #12
    I had a go at it, but as usual got nowhere.

    I want to prove that

    [tex]\langle Sx\,|\,y \rangle = \langle Tx\,|\,x \rangle \quad \Rightarrow \quad S=T[/tex]

    for all [itex]x,y \in \mathcal{H}[/itex] and for all [itex]S,T \in \mathcal{B(H)}[/itex].


    So I began by supposing that [itex]\langle Sx\,|\,y \rangle = \langle Tx\,|\,x \rangle[/itex]. I want to prove [itex]S=T[/itex] but proving [itex]S-T=0[/itex] will suffice.

    Now we know that the orthogonal complement of a Hilbert space is 0 (and [itex]S-T=0[/itex]), so it we must show that

    [tex]S=0 \quad \Leftrightarrow \langle Sx\,|\,y \rangle = 0[/tex]

    for every [itex]x,y \in \mathcal{H}[/itex].


    The polarization identity says that

    [tex]4\langle Sx\,|\y \rangle = \langle S(x+y)\,|\,(x+y) \rangle - \langle S(x+y)\,|\,(x-y) \rangle + i\langle S(x+iy)\,|\,(x+iy) \rangle - i\langle S(x-iy)\,|\,(x-iy) \rangle[/tex].
     
  14. Sep 30, 2005 #13

    George Jones

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    Let [itex]A = S - T[/itex]. Then

    [tex]
    \langle Sx\,|\,y \rangle = \langle Tx\,|\,y \rangle \Leftrightarrow \langle Ax\,|\,y \rangle = 0.
    [/tex]

    Fix [itex]x[/itex]. Since the above is true for every [itex]y[/itex], [itex]Ax=0[/itex]. But [itex]x[/itex] was arbitary, and thus [itex]Ax=0[/itex] for every [itex]x[/itex]. Consequently, [itex]A=0[/itex].

    Regards,
    George
     
  15. Sep 30, 2005 #14
    Do you need the condition whether x is not 0 since you say it is arbitrary?
     
  16. Sep 30, 2005 #15

    George Jones

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    I meant to say it was arbitrary; actually, I said it was "arbitary."

    I don't think any restriction needs to be placed on x.

    Regards,
    George
     
  17. Sep 30, 2005 #16
    But the condition in the given problem is All x belongs to H, and what if x=0 ? A doesn't need to equal to o but still satisfies the condition Ax=0, I just mean that point.
     
  18. Sep 30, 2005 #17

    George Jones

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    It's true that if Ax = 0 for every non-zero x, then A = 0, but the case x = 0 doesn't have to be excluded. If Ax = 0 for every x, then A = 0.

    Regards,
    George
     
  19. Sep 30, 2005 #18
    If so, the original problem mistook about the condition, not All x,y in H satisfy the problem.
     
  20. Sep 30, 2005 #19
    Is your formula correct ? Can you rewrite it if it were wrong since I don't have the book to check it now ?
     
  21. Sep 30, 2005 #20
    Ok, so I suppose no Polarization identity needed then.

    So we know that [itex]\langle Sx\,|\,y \rangle = \langle Tx\,|\,y \rangle[/itex]

    then rearrangining gives

    [tex]\langle Sx\,|\,y \rangle - \langle Tx\,|\,y \rangle = 0[/tex]

    which implies

    [tex]\langle (Sx-Tx)\,|\,y \rangle = 0[/tex]

    which implies

    [tex]\langle (S-T)x\,|\,y \rangle = 0[/tex]

    Let [itex]A = S-T[/itex], then

    [tex]\langle Ax\,|\,y \rangle = 0[/tex]

    Now fix [itex]x \in \mathcal{H}[/itex]. Since the above is true for every [itex]y \in \mathcal{H}[/itex], then [itex]Ax = 0[/itex] for every [itex]x[/itex]. Now if [itex]x=0[/itex] then this does not imply [itex]Ax = 0[/itex]. So for every [itex]x\neq 0[/itex], [itex]Ax=0\,\Rightarrow A=0[/itex].
     
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