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Question about Linear Operator's Image and Kernel

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data

    If [itex]T:V\rightarrow V[/itex] is linear, then Ker(T^2)=Ker(T) implies Im(T^2)=Im(T).

    2. Relevant equations

    Let [itex]T:V\rightarrow V[/itex] be a linear operator such that [itex]\forall x\in V[/itex],

    [itex]T^2(x)=0\Rightarrow T(x)=0[/itex] (Ker(T^2)=Ker(T)).​

    Prove that [itex]\forall x\in V, \exists u\in V\ni T(x)=T^2(u)[/itex] (Im(T^2)=Im(T)).

    3. The attempt at a solution

    Any clue on where I should start? I'm really stuck at this problem and have been thinking about it for the past two days. The problem is I don't know how to use the assumption Ker(T)=Ker(T^2) .
     
  2. jcsd
  3. Oct 12, 2011 #2
    Try to prove the inclusions [itex] \text{im}(T) \subseteq \text{im}(T^2)[/itex] and [itex] \text{im}(T) \supseteq \text{im}(T^2)[/itex] separately. One inclusion should be easy and the other will require slightly more work.

    There is a theorem from which this result is immediate, but you might not have seen it before...
     
  4. Oct 12, 2011 #3
    I've managed to show that [itex]Im(T^2)\subseteq Im(T)[/itex]. The other direction is my problem. But somehow, it has struck me just now that maybe I can use the rank-nullity theorem to show it. The theorem implies that [itex] dim(Im(T^2))=dim(Im(T))[/itex]. And since both of them are vector spaces, it suffices to show that two vector spaces with the same vector addition and scalar multiplication and also dimension are indeed equal.

    Furthermore, since [itex]Im(T^2)\subseteq Im(T)[/itex], it suffices to show that a subspace with the same dimension as its superspace (I don't know the correct term) is indeed equal to the latter. This leads to the question: does the set of linearly independent vectors (the basis) of the subspace span the superspace?
     
  5. Oct 12, 2011 #4
    I have found the answer to the previous question. Let A be the basis of [itex]Im(T^2)[/itex]. IF if there exists [itex]v\in Im(T)-Span(A)[/itex], then v is linearly independent from A and hence the number of linear independent vectors of [itex]Im(T^2)[/itex] is plus one than the maximum number (the number of basis vectors). This is a contradiction. Hence, [itex]Im(T)=Span(A)=Im(T^2)[/itex]. Please let me know if I've made some mistakes.
     
  6. Oct 12, 2011 #5
  7. Oct 12, 2011 #6
    Yes, I forgot to mention, V is finite dimensional. Oh, I'm not familiar with the theorem as I'm also not familiar with algebraic structures. Do you think that the problem can also be proven in the infinite dimensional case?
     
    Last edited: Oct 12, 2011
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