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Given a linear operator $K$ on $W,$ and that for a non-negative n range $K^n = $ range $K^{n+1}$ how do I prove that range $K^p = $ range $K^n$ for all $p > n?$ The first direction is trivial, the second I stumble on!

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you are trying to prove the first statement if and only if the second statement? –  smackcrane Nov 6 '11 at 23:39
    
What is $m$? The same as $n$? –  lhf Nov 6 '11 at 23:43
    
@lhf: Sorry, corrected. I am trying to prove the second statement given that I know it's a linear operator and that for nonnegative powers the equality I gave holds. –  ranges_boreicu Nov 7 '11 at 0:13

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If you know that for some $n \in \mathbb{N}$, $range K^n = range K^{n+1}$, and want to prove that for all $p \in \mathbb{N} (p>n), range K^p = range K^n$ then induction will certainly work.

To make it easier to see the argument, let $U = range K^n$; then since $range K^{n+1} = range KK^{n}$, the statement $range K^n = range K^{n+1}$ becomes $U = range K(U)$. Intuitively, we see that repeated applications of the linear operator $K$ do not change the range, so we should be able to prove the desired result.

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I have n as one single non-negative integer for which I know this holds, sorry if I miswrote. Would induction still work on given that it's a specific n we know about? –  ranges_boreicu Nov 7 '11 at 0:21
    
@ranges: Yes; the same statement as discipulus writes, except that instead of being arbitrary natural numbers, $a$ and $b$ are arbitrary natural numbers greater than or equal to $n$. That is, do "bounded induction". –  Arturo Magidin Nov 7 '11 at 0:27

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