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Prove that T is injective if and only if T* is surjective

  1. Jul 14, 2009 #1
    1. The problem statement, all variables and given/known data
    T ∈ L(V,W). Thread title.


    2. Relevant equations



    3. The attempt at a solution
    Note that T* is the adjoint operator. But there's one thing that I need to get out of the way before I even start the proof. Now consider <Tv, w>=<v, T*w> w in W, v in V. Now when they say T is injective if and only if T* is surjective, <Tv, w>=<v, T*w> only makes sense when W=V since Tv is supposed to map to W and if dim W<dim V, T is not injective.
     
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  3. Jul 14, 2009 #2

    Office_Shredder

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    I think by definition T must map a space V to itself in order to have an 'adjoint'. At least the definitions I've seen. But regardless, for <Tv,w> = <v,T*w> we only need that T* maps W to V, not that W=V. In this case the inner product on the left hand side would be the one for W, and the one on the right would be for V
     
  4. Jul 14, 2009 #3
    Oh, sorry to argue, but my textbook gave an example where T* mapped a vector
    from R2 to R3. This led me to the conclusion that
    T* maps from the subspace to the space, unless you mean V to itself as in
    a subspace of V (which is in V) back to V. I appreciate the help, btw.
     
  5. Jul 14, 2009 #4

    Office_Shredder

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    Why is T* mapping from a subspace to a space? T* is just mapping between two spaces... if V and W are different, they aren't required to be contained in each other are they?
     
  6. Jul 14, 2009 #5

    Dick

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    Pick bases for V and W. If M is the matrix representing T, then the transpose of M is the matrix representing T*. Now remember row rank=column rank. What does rank have to do with being injective or surjective?
     
    Last edited: Jul 14, 2009
  7. Jul 14, 2009 #6
    That makes life alot easier. The author never did anything with matrices with regards
    to adjoints. Rank is important because the number of columns=the dimension of the (sub)space
    and if the rank of Mv=the rank of M, the transformation is injective (and surjective).
    If the rank of Mv is less than the rank of M, the transformation is surjective (has to do with
    excess 0 rows or columns "disappearing"). Don't mind me much, haven't done matrices in a couple of weeks,
    bit rusty.
     
  8. Jul 14, 2009 #7

    Dick

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    I think so, yes, something like that. An mxn matrix is injective if it has full rank and the corresponding nxm transpose matrix is surjective if it has full rank.
     
    Last edited: Jul 14, 2009
  9. Jul 14, 2009 #8
    So in that case, MT maps a vector to the same space as Mv.
    So now I see what they mean by T* in (W, V). I mean, mapping to W is still
    mapping to V. In the end, excess rows or columns will dissappear until we have
    an nxn matrix.
     
  10. Jul 15, 2009 #9

    HallsofIvy

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    I have never seen such a definition. As you say, if T maps V to W, T* must map W to V. In order that a linear transformation to be "self-adjoint", that is that T*= T, then T must map V to itself.
     
  11. Jul 15, 2009 #10
    Oh should <Tv, w>=<v, T*w> be considered? I think Dick's transpose idea makes
    me think of the adjoint in a much easier, intuitive way. Btw I think my textbook
    did a poor job at explaining adjoints, I did some searching online, all I could find
    were explanations beyond my level of math.
     
    Last edited: Jul 15, 2009
  12. Jul 15, 2009 #11
    But I did a little thinking and that we can consider M to be a 3x4 matrix of three columns,
    4 rows and M represents the injective T. Now I can see that Mw becomes a matrix
    with four rows and three columns which represents going from W to V. But then
    the bottom row "disappears" giving a 3x3 matrix which represents what has
    actually happened. So basically, mapping from W--->V is the same as mapping from
    W--->W which is obviously an injective map. After all W is in V, right? I sound like a broken record, but this is not the actual proof, lol.
    Hold on, I can see where <Tv, w>=<v, T*w> comes into play here. I mean, I know that this is equivalent to
    <T*v, w>=<v, Tw>. So using the 3x4 M matrix as above for T, Mw becomes a 3x4 matrix with zeroes at row 4.
    We cannot multiply the 4 component v into a matrix of 3x3, but in a sense, the 3x4 matrix with zeroes at the bottom
    is the same as the 3x3 matrix.
     
    Last edited: Jul 15, 2009
  13. Jul 15, 2009 #12
    am i right here?
     
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