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Math Help - Iterated Kernels and Images

  1. #1
    Newbie Hugal's Avatar
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    Iterated Kernels and Images

    Yo !
    I've a problem about kernel and endomorphism. And i'm a little bit worried 'cause i'm blocked when i've just answered the first question. But let's see...

    Ok, so we have E a \mathbb{K}-vector space and f an endomophism of E.
    We have f^0 = \text{Id}_E and \forall k \geq 1, f^k = f \circ f^{k-1}.

    So, i've prooved that we always have, for all k \geq 0,
    \displaystyle \text{Im} f^{k+1} \subset \text{Im} f^k and
    \displaystyle \text{Ker} f^{k} \subset \text{Ker} f^{k+1}

    Now, the question is
    Show that, if \text{Ker} f^{k} = \text{Ker} f^{k+1}, then \forall m\geq k, \text{Ker} f^{m} = \text{Ker} f^{k}
    Ok, now of course i've searche for a little while. But the fact is : i can't see how to start the proof. I think of several way : to prove it directly, maybe by induction, i've even think of trying it with contraposition !

    Well, so i really don't want the answer ! Please !
    But just the smallest hint ( such as, the way to prove it ) which can make me see how to keep going.

    Thanks for reading me ! And i apologize if my english is poor

    Hugal.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    The key is to prove:

    \ker f^{k+1}\subset \ker f^k\Rightarrow \ker f^{k+2}\subset \ker f^k


    In fact,


    x\in \ker f^{k+2} \Rightarrow f^{k+2}(x)=f^{k+1}(f(x))=0\Rightarrow

    f(x)\in \ker f^{k+1}\subset \ker f^k\\\Rightarrow f^k(f(x))=0 \Rightarrow

    f^{k+1}(x)=0 \Rightarrow x\in \ker f^{k+1}\subset \ker f^k


    Fernando Revilla
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  3. #3
    Newbie Hugal's Avatar
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    Thanks a lot,
    It was really easy... i'm just stupid :P

    Thanks again.
    Hugal.
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  4. #4
    Newbie Hugal's Avatar
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    Ok, now i really feel stupid since i've been found blocked again a few question further :P
    Here it is :
    Let (assuming they exist)
     s = \min\{ k \geq 1, \text{ker} f^k = \text{ker} f^{k+1} \}
    and  r = \min\{ k \geq 1, \text{Im} f^k = \text{Im} f^{k+1} \}

    Proove that, if r \leq s, then  \text{ker} f^s = \text{ker} f^{r}
    Of course, i've tried to begin, as follow :

    We already know that \text{ker} f^r \subset \text{ker} f^{s} (since r \leq s).
    So we just have to show that \text{ker} f^s \subset \text{ker} f^{r}.
    So,
    x \in \text{ker} f^s \Rightarrow f^s (x) = 0

    From now on, i'm not quite sure. We may expand f^s as f^r \circ f^{s-r} and then using the fact that \text{ker} f^r \subset \text{ker} f^{s}.
    But it doesn't seems to end up very well :/

    So, i just want a small hint to help me progress, not the whole answer !
    Thanks again for reading and, eventually, helping me.

    Hugal.
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Hugal View Post
    So, i just want a small hint to help me progress, not the whole answer !
    We have already proved:

     \ker f^k=\ker f^{k+1}=\ker f^{k+2}

    then,

    \ker f^{p+1}=\ker f^p,\;(p=k+1)

    \dots


    Fernando Revilla
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  6. #6
    Newbie Hugal's Avatar
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    Well, i've meditated on that message, but i am not quite sure i've really understood.

    Ok, so i try to sum all up.

    We have prooved that :

    \displaystyle \forall k \geq s, \text{ker} f^k = \text{ker} f^s

    Or,

    \displaystyle \forall k \geq s, \text{ker} f^s = \text{ker} f^{k} = \text{ker} f^{k+1} = \text{ker} f^{k+2} = \text{ker} f^{k+3}

    But what you just changed is the name of the variable, so we may write, for example

    \displaystyle\text{ker} f^s = \text{ker} f^{k} = \text{ker} f^{p} = \text{ker} f^{q} = \text{ker} f^{t}

    and i still don't see where it leads us, for we just change some names :/
    Moreover, we won't be able to get back before s with such equalities because r \leq s.

    So, i apologize, but i did not understand very well what you meant.

    Hugal.
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  7. #7
    MHF Contributor FernandoRevilla's Avatar
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    (i) You have proved

    \ker f^k\subset \ker f^{k+1},\;\forall k\geq 0

    (ii) Now, we have to prove

      \ker f^{k+1}=\ker f^k\Rightarrow \ker f^m=\ker f^k,\forall m\geq k\quad (*)

    (iii) (*) is equivalent to

    S(p):\;\; \ker f^{k+1}\subset \ker f^k\Rightarrow \ker f^{k+p}\subset \ker f^k,\forall p\geq 0\quad [1]

    (iv) S(1) is trivial (by hyphothesis) , and S(2) was proved in a previous post.

    (v) Now, complete the induction method.


    Fernando Revilla
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  8. #8
    Newbie Hugal's Avatar
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    Ok, now i'm disturbed.
    I'm afraid to say something stupid, but i think there have been a misunderstanding. :/
    Because what you make me proof here is the answer to the first question i ask, nope ?
    Your last post prooved what i've ask in my very first post :

    Show that, if \text{ker} f^k = \text{ker} f^{k+1}, then \forall m \geq k, \text{ker} f^m = \text{ker} f^k
    didn't it ? :/

    Now i'm kind of embarrassed because we've gotta prove it for a integer r \leq k so this induction won't work...

    Hoping i didn't say something completely stupid :s

    Hugal.
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  9. #9
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Hugal View Post
    Because what you make me proof here is the answer to the first question i ask, nope ?

    You quoted the new question so, I was not aware of it. Let us see, if:

    s=\min\{k\geq 1:\ker f^k=\ker f^{k+1}\}

    then,

    \ldots \subsetneq \ker f^{s-2}\subsetneq \ker f^{s-1}\subsetneq \ker f^{s}=\ker f^{s+1}=\ker f^{s+2}=\ldots

    and independently on the way you define r\leq s we have \ker f^s\neq \ker f^r if r<s and \ker f^s=\ker f^r if r=s .
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  10. #10
    Newbie Hugal's Avatar
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    Ok, now i think i got it.
    Maybe it is the way my question is asked which is weird, because it is exactly :

    "Proove that, if r \leq s, then \ker f^s = \ker f^r. Then s = r"

    So, i've tried something desperate with a lot of complicated things and ended up with... well nothing.
    But i think now it is clear.

    My excuses for the misunderstanding and my thanking for the help

    Hugal.
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