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Rank nullity inequality fom T^2

  1. Mar 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Let T be a linear operator on a vector space V (finite dimensional). show that
    nullity(T^2)<= 2nullity(T)

    2. Relevant equations
    the rank nullity theorem.


    3. The attempt at a solution
    i take T to be a lin. transf. from ker(T^2) to V, that is i am restricting the domain of T.

    well, i show kerT is a subset of ker(T^2) then i write
    nullity(T^2)= nullity(T(Ker(T^2)) + dimT(ker(T^2))= dimkerT +dimT(ker(T^2))

    and this last equality i am not sure of. then i show T(kerT^2) is a subset of kerT and the theorem follows from the equality which i am not sure of. the proff. said to use the fact that kerT is a subset of ker(T^2). but i don't see how this helps. i started by examining T(kerT^2) and said that oh... okay i get it now. since kerT is a subset of kerT^2 there could be some vectors ouside of the kerT in this latter set. but that still makes the kernel of T(kerT^2) equal to the kerT since it is the kernel of {0, T(v1), ... T(vn)}. is this correct, please respond i have wasted so much time on this damn thing.
     
  2. jcsd
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