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Basis for the image of a surjective linear map.

  1. Jun 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Let V and W be vector spaces over F, and let T: V -> W be a surjective (onto) linear map. Suppose that {v1, ..., v_m, u1, ... , u_n} is a basis for V such that ker(T) = span({u1, ... , u_n}). Show that {T(v1), ... , T(v_m)} is a basis for W.


    2. Relevant equations
    Basic properties of linear maps. Linear independence.


    3. The attempt at a solution
    I have already proven that {T(v1), ... , T(v_m)} spans W, which I thought would be harder than showing linear independence. But here is where I am confused. We have to show that {T(v1), ... , T(v_m)} is linearly independent.

    Suppose (a_1)T(v1) + ... + (a_m)T(v_m) = 0 for (a_1), ... (a_m) in F. Then
    T( (a_1)(v1) + ... + (a_m)(v_m) ) = 0 since T is linear. Now if we also suppose that (a_1)(v1) + ... + (a_m)(v_m) = 0, then clearly (a_1) = (a_2) = ... = (a_m) = 0 since the set {v_1, ... , v_m} is linearly independent.

    But I think I'm confused when (a_1)(v1) + ... + (a_m)(v_m) =/= 0 (which is certainly possible right?). However, I have an idea and I think that in this case, we still get (a_1) = ... = (a_m) = 0?
     
  2. jcsd
  3. Jun 5, 2009 #2
    well, if T(w)=0 then w is in the KerT, so we can write this as a linear combination of the u's.

    you shouldn't need any more of a hint than this.
     
  4. Jun 5, 2009 #3
    Thanks xao, I think that was exactly what I had in mind (and I should have sorted out this case before I made the post). Basically then, setting a linear combination of the v's equal to a linear combination of the u's implies (by isolating 0 on one side) that all coefficients of vectors are 0.
     
  5. Jun 5, 2009 #4

    HallsofIvy

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    Sorry, I misread the problem.
     
    Last edited: Jun 7, 2009
  6. Jun 5, 2009 #5
    But ker(T) = {0} if and only if T is injective, and surjectivity does not rule out the possibility that a nonzero vector is mapped to the 0 vector.

    And just to clarify, the u's xao is talking about are also in the linearly independent set (consisting of v1 to v_m and u1 to u_n, sorry for not using latex). So currently my proof of linear independence looks like this:

    Suppose

    [itex]
    a_1T(v_1)+ a_2T(v_2)+ \cdot\cdot\cdot+ a_mT(v_m)= T(a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_mv_m)= 0[/itex]

    But kert(T) = span({u1, ... , u_n}), so we can write
    [itex]a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_mv_m = b_1u_1+ b_2v_2+ \cdot\cdot\cdot+ b_nu_n.[/itex]

    Then
    [itex]a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_mv_m = b_1u_1 - b_2v_2 - \cdot\cdot\cdot - b_nu_n = 0[/itex]

    and it follows that [itex]a_1 = a_2 = \cdot\cdot\cdot= a_m = 0[/itex].
     
    Last edited: Jun 5, 2009
  7. Jun 6, 2009 #6
    i'm hoping that '=' was a typo in the last step.

    it looks like the dimension of V is m+n (the v's plus u's) while the dimension of W is m, so a surjective map only needs m (the v's) of the m+n basis vectors. that leaves n basis vectors (the u's) spanning KerT to map onto the zero of W.
     
  8. Jun 7, 2009 #7
    Yeah sorry I really screwed up the latex on that last line. It should read

    Then

    [itex]
    a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_mv_m - b_1u_1 - b_2u_2 - \cdot\cdot\cdot - b_nu_n = 0
    [/itex]

    so the conclusion follows.
     
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