Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Image and kernel of T^n

  1. Jan 12, 2005 #1
    Hi,

    What's the relationship between the image and kernel of T and the image and kernel of Tn? I think we saw in class something along the lines of:

    [tex]Ker(T) \subseteq Ker(T^2)[/tex]
    [tex]Im(T) \supseteq Im(T^2)[/tex]

    My intuition is that this is also correct for any natural n, but is it true and if so how do you prove it, by induction?

    Thanks,
    Chen
     
  2. jcsd
  3. Jan 12, 2005 #2
    Assuming that T is a linear map from some vector space V to itself... [tex]\ker{T} \subseteq \ker{T^n}[/tex] (where n is any natural number) is easy to prove. It's obvious for n = 1.

    Suppose it's true for n = k. If [itex]x \in \ker{T}[/itex], then [itex]x \in \ker{T^k}[/itex], by the induction hypothesis. But then [itex]T^{k+1}(x) = T(T^{k}(x)) = T(0) = 0[/itex] (by the linearity of T), so that [itex]x \in \ker{T^{k+1}}[/itex].

    I think the thing about Im(T) could be done without induction.
     
    Last edited: Jan 12, 2005
  4. Jan 12, 2005 #3
    Let me clear the cobwebs from out my skull...
    Let T be a linear map from V to V'.

    A_1=Ker(T)={v in V : Tv=0}.

    A_2=Ker(T^2)={v in V : TTv=0}.

    Note that T0=0 for all linear maps T. Here, the first 0 is in V and the second 0 is in V'.

    Then A_1 is a subset of A_2 because:
    v in A_1 implies
    Tv=0 implies
    TTv=T0=0
    implies v in A_2.

    I don't think you really need induction if you can get away with saying that (T^n)0=0. I suppose that technically you do need induction if it's not acceptable as being obvious: T0=0, so done when n=1. Then assuming (T^(n-1))0=0, we can apply T to both sides to get (T^n)0=T0=0. Done.

    Then if you let A_n=Ker(T^n), go through the above proof to show that A_(n-1) is a subset of A_n (change 1 to n-1 and 2 to n).

    Then you have the following result:
    A_1 is a subset of A_2 is a subset of ... is a subset of A_n.

    The image I'll work out if no one else does after I have a cigarrette...
     
  5. Jan 14, 2005 #4

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    Note that for a linear operator T on a vector space V, we have

    [tex]Im(T^2) = T(T(V)) = T(Im(T))[/tex]

    whereas [itex]Im(T) = T(V)[/itex]. Clearly, since [itex]Im(T) \subseteq V[/itex], [itex]T(Im(T)) \subseteq T(V)[/itex] follows immediately. For any n, we can compare [itex]Im(T^n)[/itex] with [itex]Im(T^{n+1})[/itex].

    [tex]Im(T^{n+1}) = T^{n+1}(V) = T^n(T(V)) = T^n(Im(T))[/tex]

    whereas [itex]Im(T^n) = T^n(V)[/itex]. Again, since [itex]Im(T) \subseteq V[/itex], it follows immediately that [itex]T^n(Im(T)) \subseteq T^n(V)[/itex], giving [itex]Im(T^{n+1}) \subseteq Im(T^n)[/itex]. This gives:

    [tex]Im(T) \subseteq Im(T^2) \subseteq \dots \subseteq Im(T^n) \subseteq \dots[/tex]

    which is what you wanted, I suppose.
     
    Last edited: Jan 14, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Image and kernel of T^n
  1. Kernels and Images (Replies: 7)

  2. Kernels and Images (Replies: 9)

Loading...