# Vector space and mapping

1. Feb 17, 2010

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. Feb 23, 2010

### Tinyboss

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.