1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook