0
$\begingroup$

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.

3 Answers 3

0

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

0

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 $m$?

0

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). $

  • 0
    Not if $f$ is not nilpotent, as you said.2012-12-13