Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

up vote 0 down vote accepted

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

share|improve this answer
    
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 $m$?

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.