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Another proof on self adjointness

  1. Jul 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that there does not exist a self-adjoint operator T ∈ L(R3)
    such that T(1, 2, 3) = (0, 0, 0) and T(2, 5, 7) = (2, 5, 7).
    2. Relevant equations



    3. The attempt at a solution
    I'm having trouble seeing that there is an actual operator, self adjoint or not,
    that can do this. I mean, according to this, T maps from R^3 to R^3
    but because it maps to R^3 and not to any subspace of R^3 (since it can map (2 5 7 ) to itself), its nullspace must be dim 0. The thing is, since (1 2 3) cannot be in the nullspace
    of T, T must be a zero transformation or its matrix might have [1 1 -1] for all three rows. But that can't happen if T(2 5 7)=(2 5 7). I mean, if T cannot exist, then neither can its adjoint.
     
  2. jcsd
  3. Jul 30, 2009 #2


    I'm not quite sure what you mean by this. Just because it maps to R3 doesn't mean that it has to map onto R3. Or maybe you mean something else, but still, knowing that one vector gets mapped to itself doesn't mean that no other vector can be mapped to zero (unless I'm missing something).

    Still, despite your misgivings, the proof is pretty straight-forward - all it takes is using T's identity as self-adjoint operator in conjunction with the properties of the inner product.
     
  4. Jul 30, 2009 #3
    Yeah thats what I meant to say: it maps onto R^3 and not some 2 or 1 or 0 dimensional
    subspace of R^3 in R^3 since T(257)=(257). Just started to get some coffee in me,
    so my wordings were a bit messed up. I'm working on this right now.
     
  5. Jul 30, 2009 #4
    Why not? Is your argument that if, for some v, Tv=v, then T must map onto R3? I'm not sure I see that.

    In our case, v = (2,5,7) and Tv=v, but all this gives us is that the one-dimensional subspace spanned by v (namely all vectors of the form av, for some scalar a) gets mapped to 0 (and Tv=0 only when a=0). We don't know anything about vectors outside of this subspace. It would not be a contradiction if dim(null(T)) were not zero.
     
  6. Jul 30, 2009 #5
    Ok <T(1 2 3), (2 5 7)>=0. For <(1 2 3), T(2 5 7)> to=<T(1 2 3), (2 5 7)>,
    <(1 2 3), T(2 5 7)> must=0. But <(1 2 3), T(2 5 7)>=<(1 2 3), (2 5 7)>.
    So T is not self adjoint since <T(1 2 3), (2 5 7)>=/=<(1 2 3 ), T(2 5 7)>.
     
  7. Jul 30, 2009 #6
    Yup, that's dead-on with the proof (though you might want to throw the word 'positive-definite' in there somewhere for clarity). Though I'm still not sure that no such operator exists :smile:
     
  8. Jul 30, 2009 #7
    Yeah, what I don't understand is when you said "In our case, v = (2,5,7) and Tv=v, but all this gives us is that the one-dimensional subspace spanned by v (namely all vectors of the form av, for some scalar a) gets mapped to 0". T(2 5 7) doesn't get mapped to zero and
    a=1.
     
  9. Jul 30, 2009 #8
    I was just trying to make sure I understood your argument and then state what I thought could be drawn from your premises. You say

    So your argument is that because T(2,5,7)=(2,5,7), the map is onto R^3, right?

    I'm just saying that all you have shown is that Tv=v for one single v (namely v=(2,5,7)), and I don't see how that is a strong enough premise to conclude that T maps onto R^3. Obviously, with v=(2,5,7) and some scalar a, T(av)=aT(v)=av, but v does not span all of R^3. You have no idea what happens to other vectors that are not multiples of v. It could easily be the case that other vectors - that is to say, those not of the form av for some a - could be mapped to zero, so a large number of vectors would be without preimages under T (slightly awkward terminology for linear algebra, but hopefully you see what I mean).
     
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