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Making T self adjoint

  1. Aug 3, 2009 #1
    Suppose U is a finite-dimensional real vector space and T ∈
    L(U). Prove that U has a basis consisting of eigenvectors of T if
    and only if there is an inner product on U that makes T into a
    self-adjoint operator.

    The question is, what exactly do they mean by "makes T into a self adjoint
    operator?" Is it that there exists an inner product of eigenvectors of T
    say <v, v> that allows T to be self adjoint?
     
  2. jcsd
  3. Aug 3, 2009 #2

    HallsofIvy

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    A linear transformation, T, on an inner product space, V, is "self adjoint" if and only if <Tu, V>= <u, TV> for all u and v in V and "< , >" is the inner product defined for V. In other words, the concept of "self adjoint" is only defined for a specific inner product.

    An inner product space is, of course, a vector space, V, for which there is a specific inner product defined. In any finite dimensional vector space, every choice of basis defines an inner product: with basis [itex]\{e_1, e_2, \cdot\cdot\cdot, e_n\}[/itex], given vectors u and v, we can write [itex]u= a_1e_1+ a_2e_2+ \cdot\cdot\cdot+ a_ne_n[/itex] and [itex]v= b_1e_1+ b_2e_2+ \cdot\cdot\cdot+ b_ne_n[/itex] and then define [itex]<u, v>= a_1b_1+ a_2b_2+ \cdot\cdot\cdot+ a_nb_n[/itex].

    Since this is an "if and only if" statement, you need to do two things:
    1) Show that if T is self adjoint under inner product < , >, then there exist basis for the vector space consisting entirely of eigenvectors of T.

    2) Show that if there exist a basis for the vector space, consisting entirely of eigenvectors of T, then T is self adjoint using the inner product defined, as above, by that basis.
     
    Last edited: Aug 4, 2009
  4. Aug 3, 2009 #3
    Ok I will use everything you have defined for a vector space, vectors, and whatnot.
    Let dimV=n.
    Now if T is self adjoint, we know that <Tv, u>=<v, Tu>. If T is self adjoint,
    then T's matrix must be nxn. Therefore T is diagonizable. This would allow T to have j-distinct eigenvalues along
    its diagonal. So there is some basis for T which contains eigenvectors, one for each eigenvalue.
     
    Last edited: Aug 3, 2009
  5. Aug 3, 2009 #4
    Now for the other direction. If there exists a basis for U consisting of eigenvectors of some T,
    then Te_i=c_ie_i. Choosing two eigenvectors from the basis we have <e_i, e_k>.
    So <Te_i, e_k>=<c_ie_i, e_k>=c_i<e_i, e_k>=<e_i, c_ie_k>=<e_i, Te_k>.
     
  6. Aug 3, 2009 #5

    Office_Shredder

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    You haven't stated what <,> is here! And ciek is not Tek in general.

    As always, it's better if you can avoid describing T as a matrix for your proof if at all possible. If it's necessary, you still have to justify everything. Skipping from "If T is self adjoint, then T's matrix must be nxn" which is a meaningless conclusion since we already know T maps a vector space to itself, to "Therefore T is diagonalizable". Why?
     
  7. Aug 3, 2009 #6
    I don't know why I said ciek=Tek, that was pretty dumb. Just got back from a long night at work.
    This will prove 2.
    Set u=a1e1+...+anen and v=b1e1+...+bnen where ek is an eigenvector.
    For the inner product <u, v> we have a1b1+...+anbn using the example from above.
    Now T has eigenvalues where Tek=ckek. Now <Tu, v>=<T(a1e1), v>+...+<T(anen), v> =c1a1b1+...+cnanbn <u, Tv>=<u, c1b1e1>+...+<u, cnbnen>=a1c1b1+...+ancnbn=c1a1b1+...+cnanbn=<Tu, v>
     
    Last edited: Aug 4, 2009
  8. Aug 4, 2009 #7
    For "1" we know T is self adjoint so <Tv, u>=<u, Tv>. But should
    I just assume that U is invariant under T, and claim that T(v)=c1b1e1+...+cnbnen
    and that bkek is in some nullspace of T-ck?. I mean we know that U is invariant
    under T and we know that T(bkek)=ck(bkek) so shouldn't that lead to ( T-ck)ek=0?
     
  9. Aug 4, 2009 #8

    HallsofIvy

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    Did you really mean to say that every "nxn" matrix is diagonalizable? That is certainly NOT true!

    Nothing you have said here makes any sense!
     
  10. Aug 4, 2009 #9
    I know. Office Shredder told me that. I fixed both parts after his post. Did you
    see the two posts before yours?
     
    Last edited: Aug 4, 2009
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