Image of power of linear operator.

How can we show that if $f:V\to V$ Then for each $m\in \mathbb {N}$ $$\operatorname{im}(f^{m+1})\subset \operatorname{im}(f^m)$$ Please help,I am stuck on this.

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Suppose that $x\in\text{im}(f^{m+1})$. This means that $x=f^{m+1}(y)$ for some $y\in V$. Then $$x=f^{m+1}(y)=f^m(f(y))\in\text{im}(f^m).$$

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How can I show that at some point it is equality i.e the subspaces equal? – p.s Dec 13 '12 at 21:13
It is not true in general. – Martin Argerami Dec 13 '12 at 21:15
I have a condition that f is not nilpotent and linear operator – p.s Dec 13 '12 at 21:17
Still not true. Take the shift operator on a Hilbert space, for example. – Martin Argerami Dec 13 '12 at 21:18
V is finite dimensional K.vector space. – p.s Dec 13 '12 at 21:23

Try it first with $m=1$. A typical member of $\def\im{\operatorname{im}} \im(f^2)$ looks like $f(f(x))$, right? Now can you see why that is a member of $\im(f)$? Next, repeat with $m=2$. Do you see how that generalizes to higher $m$?

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For $m\in \mathbb{N}$ : $$f^m(f(V))\subset imf^m$$ since $f(V)\subset V$, i.e $$imf^{m+1}\subset imf^m$$.

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