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Vector space and mapping

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data
    let T(V)=V be a linear map, where V is a finite-dimensional vector space. Then T^2 is defined to be the composite TT of T with itself, and similarly T^(i+1) = TT^i for all i >=1. Suppose Rank (T) = Rank (T^2)

    2. Relevant equations

    a) prove that Im(T) = Im(T^2)

    b) for i>=1, let U_i(Im(T))=Im(T) be definied as the restriction of T^i to the subspace Im(T) of V. Show that U_i is nonsingular for all i

    c) Deduce that Rank (T) = Rank (T^i) for all i >= 1

    3. The attempt at a solution

    a) since Rank (T) = Rank (T^2), then
    dim( Im(T) ) = dim ( Im(T^2) )
    since Im (T) = V and TT(V) = T(V) = V = Im(T^2)
    so Im(T) = Im (T^2) because V=V
    -does that make sense?-

    b) U_i (Im(T)) = Im(T)
    so U_i (V) = V
    since V is mapped to itself, U_i has to be an identity matrix
    and identity matrix has an inverse because its determinant is not zero,
    so U_i is nonsingular
    -does that make sense?-

    c) since T is a linear map that V to itself, T^i (V) = V for all i>=1,
    implies that image would be the same
    hence dimension of image is the same
    so the Rank (T) = Rank (T^i) for all i>=1
    -again, does that make sense?-
  2. jcsd
  3. Feb 23, 2010 #2
    b) Just because T(V)=V does not mean T is the identity. It doesn't mean "Tv=v for all v in V". It just means it's a surjective map. However, you can conclude that T is invertible.

    c) Right idea, but you should probably use induction explicitly.
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