kernels of powers of linear maps Say that $V$ is a finite dimensional vector space over a field and and $f : V \to V$ a linear map. There is an integer $i$ such that $\text{ker}(f^n) = \text{ker}(f^{n+1})$ for all $n \geq i$. You see that by noting that $\text{ker}(f^n) \subseteq \text{ker}(f^{n+1})$ for all $n$ and since $V$ is finite dimensional, they must stabilize at some point.
I am having trouble seeing that for $n \leq s$ where $s$ is the least integer $i$ above, that $\text{ker}(f^n) \subsetneq \text{ker}(f^{n+1})$. How can I see the proper containment?
I am pretty sure that the proof goes like this: If $n < s$ and $\text{ker}(f^{n-1}) = \text{ker}(f^{n})$, then $\text{ker}(f^j) = \text{ker}(f^{j+1})$ for all $j \geq n$ contradicting that $s$ is the least such integer. But how do I show this? Thanks for your help.
 A: Let $K_i=\ker(f^i)$. Assume, that $n\in\mathbb{Z}_{\geq 0}$ and $K_n=K_{n+1}$. We show that $K_{n+1}=K_{n+2}$. Clearly, this will prove your claim. Obviously, $K_{n+1}\subset K_{n+2}$. Conversely, let $a\in K_{n+2}$ and let's show that $a\in K_{n+1}$. Since $a\in K_{n+2}$, then 
$0=f^{n+2}(a)=f^{n+1}(f(a))$, hence $f(a)\in K_{n+1}=K_n$. Therefore $0=f^n(f(a))=f^{n+1}(a)$, so $a\in K_{n+1}$.
A: You are right about your proof. How to do it ? 
It's only using rank–nullity theorem many times.
I take your notation :
If $ker(f^{n-1})=ker(f^n)$, using rank-nullity theorem : $$dim(im(f^{n}))=dim(V)-dim(ker(f^n))=dim(V)-dim(ker(f^{n-1}))=dim(im(f^{n-1}))$$and the fact that $im(f^{n})$ is included in $im(f^{n-1})$ it shows that $im(f^{n-1})=im(f^{n})$.
Then knowing that $im(f^{n+1})$ is included in $im(f^{n})$, let $x$ be in $im(f^{n})$. 
Then there exists $y\in V$ such that $x=f^n(y)=f(f^{n-1}(y))$ and $f^{n-1}(y)\in im(f^{n-1})=im(f^{n})$ so there exists $z\in V$ such that $f^{n-1}(y)=f^n(z)$. Then $x=f(f^n(z))\in im(f^{n+1})$. 
Having both inclusions : $im(f^{n+1})=im(f^{n})$. Now using rank-nullity theorem : by inclusion and dimension equality  $$\ker(f^{n+1})=\ker(f^{n}).$$
This is the initialization of the induction and in the exact same way you can show the equality of the kernels until and reaching $s$.
Am I clear? I can detail more.
