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All eigenvalues zero => zero map

  1. Sep 29, 2007 #1
    I want to prove that if all the eigenvalues of a linear transformation T : V --> V are zero, then T = 0. I think this is obvious but I'm having difficulty putting it into words.

    If all the eigenvalues of T are zero, then there exists a basis B for V in which [itex][T]_B [/itex] is the zero matrix. Thus [itex][T(v)]_B=[T]^B_B[v]_B = 0\; \forall v \in V[/itex].

    I think I should show that the matrix representation of T in an arbitrary basis D is zero. We have from second year linear algebra [itex][id]^B_D [T]^B_B [id]^D_B = [T]^D_D[/itex].
    Thus

    [itex][T(v)]_D = [T]^D_D [v]_D = [id]^B_D [T]^B_B [id]^D_B [v]_B = 0[v]_D = 0[/itex].

    But [itex]v \in V[/itex] is abritrary so [itex][T]^D_D [v]_D = 0 \; \forall [v]_D \in \mathbb{F}^n[/itex] where [itex]\mathbb{F}[/itex] is the base field. Hence T = 0.

    Is this correct?
     
    Last edited: Sep 29, 2007
  2. jcsd
  3. Sep 29, 2007 #2

    Hurkyl

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    This is not a theorem! There are at least two ways this can fail:

    This matrix over the reals has only one eigenvalue, 0. (It has 3 over the complexes, of course)
    [tex]
    \left(
    \begin{array}{ccc}
    0 & 0 & 0 \\
    0 & 0 & -1 \\
    0 & 1 & 0
    \end{array}
    \right)
    [/tex]

    This matrix over any field has only one eigenvalue, 0, of (algebraic) multiplicity 2.
    [tex]
    \left(
    \begin{array}{cc}
    0 & 1 \\
    0 & 0
    \end{array}
    \right)
    [/tex]
     
    Last edited: Sep 29, 2007
  4. Sep 29, 2007 #3

    Hurkyl

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    By the way, you don't have to deal with that "arbitrary basis" stuff.

    The zero linear transformation from V to W is the linear transformation such that, for every v in V,
    Tv = 0.​

    If you can find a basis so that the coordinate representation of T is the zero matrix, then it's easy to show that T is the zero transformation.

    (Recall that two functions f and g are equal if and only if f(x) = g(x) for every input x)
     
    Last edited: Sep 29, 2007
  5. Sep 29, 2007 #4

    Hurkyl

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    That doesn't look right: [itex][v]_D[/itex] and [itex][v]_B[/itex] are usually different.

    Oh, but I think it's just a typo, because you switched back to [itex][v]_D[/itex] in the next expression.


    You could have done the basis calculation in a shorter way:
    [tex]
    [T(v)]_D = [id]_D^B [T(v)]_B = [id]_D^B 0 = 0.
    [/tex]
    Other than your typo, I think you did the basis manipulations correctly.
     
    Last edited: Sep 29, 2007
  6. Sep 29, 2007 #5
    Hi Hurkyl,

    You're right, my theorem is a little bit too general.

    Let T be an Hermitian linear operator.

    Now I think it's okay.

    And yes, that's a typo.

    Btw, I don't see any other "easy" way to the proof based on what you said about equal functions.
     
  7. Sep 29, 2007 #6

    Hurkyl

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    (caveat: this is only for finite-dimensional linear algebra. I'm not sure what generalizes properly to infinite dimensions)


    That sounds fine; Hermitian matrices are diagonalizable over the reals, and I believe that is sufficient to prove your theorem. (BTW, you can do the proof by working directly with the diagonalized form!)
     
  8. Sep 29, 2007 #7

    Hurkyl

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    Starting from
    For every v, [itex][T(v)]_B = 0 = [0]_B[/itex]​
    it follows that
    For every v, T(v) = 0
    and thus
    T = 0.​
     
  9. Sep 29, 2007 #8
    Yeah, but to get from assumption to therefore you need what I said right? Like l said it's obvious.
     
  10. Sep 29, 2007 #9

    Hurkyl

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    It follows simply because the coordinate representation is faithful:
    [tex]v = w \text{\ if and only if\ } [v]_B = [w]_B.[/tex]
    Of course, your way is correct too.
     
  11. Sep 29, 2007 #10

    Hurkyl

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    *bonks self* I just realized this is exactly what you are doing when you said you could find a basis relative to which T's coordiante representation is the zero matrix!
     
  12. Sep 29, 2007 #11
    Right, like saying [itex][\cdot]_B: V \to \mathbb{F}^n[/itex] is injective. Thanks.
     
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