1. Iterated Kernels and Images

Yo !
I've a problem about kernel and endomorphism. And i'm a little bit worried 'cause i'm blocked when i've just answered the first question. But let's see...

Ok, so we have $\displaystyle E$ a $\displaystyle \mathbb{K}$-vector space and $\displaystyle f$ an endomophism of $\displaystyle E$.
We have $\displaystyle f^0 = \text{Id}_E$ and $\displaystyle \forall k \geq 1, f^k = f \circ f^{k-1}$.

So, i've prooved that we always have, for all $\displaystyle k \geq 0$,
$\displaystyle \displaystyle \text{Im} f^{k+1} \subset \text{Im} f^k$ and
$\displaystyle \displaystyle \text{Ker} f^{k} \subset \text{Ker} f^{k+1}$

Now, the question is
Show that, if $\displaystyle \text{Ker} f^{k} = \text{Ker} f^{k+1}$, then $\displaystyle \forall m\geq k, \text{Ker} f^{m} = \text{Ker} f^{k}$
Ok, now of course i've searche for a little while. But the fact is : i can't see how to start the proof. I think of several way : to prove it directly, maybe by induction, i've even think of trying it with contraposition !

But just the smallest hint ( such as, the way to prove it ) which can make me see how to keep going.

Thanks for reading me ! And i apologize if my english is poor

Hugal.

2. The key is to prove:

$\displaystyle \ker f^{k+1}\subset \ker f^k\Rightarrow \ker f^{k+2}\subset \ker f^k$

In fact,

$\displaystyle x\in \ker f^{k+2} \Rightarrow f^{k+2}(x)=f^{k+1}(f(x))=0\Rightarrow$

$\displaystyle f(x)\in \ker f^{k+1}\subset \ker f^k\\\Rightarrow f^k(f(x))=0 \Rightarrow$

$\displaystyle f^{k+1}(x)=0 \Rightarrow x\in \ker f^{k+1}\subset \ker f^k$

Fernando Revilla

3. Thanks a lot,
It was really easy... i'm just stupid :P

Thanks again.
Hugal.

4. Ok, now i really feel stupid since i've been found blocked again a few question further :P
Here it is :
Let (assuming they exist)
$\displaystyle s = \min\{ k \geq 1, \text{ker} f^k = \text{ker} f^{k+1} \}$
and $\displaystyle r = \min\{ k \geq 1, \text{Im} f^k = \text{Im} f^{k+1} \}$

Proove that, if $\displaystyle r \leq s$, then $\displaystyle \text{ker} f^s = \text{ker} f^{r}$
Of course, i've tried to begin, as follow :

We already know that $\displaystyle \text{ker} f^r \subset \text{ker} f^{s}$ (since $\displaystyle r \leq s$).
So we just have to show that $\displaystyle \text{ker} f^s \subset \text{ker} f^{r}$.
So,
$\displaystyle x \in \text{ker} f^s \Rightarrow f^s (x) = 0$

From now on, i'm not quite sure. We may expand $\displaystyle f^s$ as $\displaystyle f^r \circ f^{s-r}$ and then using the fact that $\displaystyle \text{ker} f^r \subset \text{ker} f^{s}$.
But it doesn't seems to end up very well :/

So, i just want a small hint to help me progress, not the whole answer !
Thanks again for reading and, eventually, helping me.

Hugal.

5. Originally Posted by Hugal
So, i just want a small hint to help me progress, not the whole answer !

$\displaystyle \ker f^k=\ker f^{k+1}=\ker f^{k+2}$

then,

$\displaystyle \ker f^{p+1}=\ker f^p,\;(p=k+1)$

$\displaystyle \dots$

Fernando Revilla

6. Well, i've meditated on that message, but i am not quite sure i've really understood.

Ok, so i try to sum all up.

We have prooved that :

$\displaystyle \displaystyle \forall k \geq s, \text{ker} f^k = \text{ker} f^s$

Or,

$\displaystyle \displaystyle \forall k \geq s, \text{ker} f^s = \text{ker} f^{k} = \text{ker} f^{k+1} = \text{ker} f^{k+2} = \text{ker} f^{k+3}$

But what you just changed is the name of the variable, so we may write, for example

$\displaystyle \displaystyle\text{ker} f^s = \text{ker} f^{k} = \text{ker} f^{p} = \text{ker} f^{q} = \text{ker} f^{t}$

and i still don't see where it leads us, for we just change some names :/
Moreover, we won't be able to get back before $\displaystyle s$ with such equalities because $\displaystyle r \leq s$.

So, i apologize, but i did not understand very well what you meant.

Hugal.

7. (i) You have proved

$\displaystyle \ker f^k\subset \ker f^{k+1},\;\forall k\geq 0$

(ii) Now, we have to prove

$\displaystyle \ker f^{k+1}=\ker f^k\Rightarrow \ker f^m=\ker f^k,\forall m\geq k\quad (*)$

(iii) $\displaystyle (*)$ is equivalent to

$\displaystyle S(p):\;\; \ker f^{k+1}\subset \ker f^k\Rightarrow \ker f^{k+p}\subset \ker f^k,\forall p\geq 0\quad [1]$

(iv) $\displaystyle S(1)$ is trivial (by hyphothesis) , and $\displaystyle S(2)$ was proved in a previous post.

(v) Now, complete the induction method.

Fernando Revilla

8. Ok, now i'm disturbed.
I'm afraid to say something stupid, but i think there have been a misunderstanding. :/
Because what you make me proof here is the answer to the first question i ask, nope ?
Your last post prooved what i've ask in my very first post :

Show that, if $\displaystyle \text{ker} f^k = \text{ker} f^{k+1}$, then $\displaystyle \forall m \geq k, \text{ker} f^m = \text{ker} f^k$
didn't it ? :/

Now i'm kind of embarrassed because we've gotta prove it for a integer $\displaystyle r \leq k$ so this induction won't work...

Hoping i didn't say something completely stupid :s

Hugal.

9. Originally Posted by Hugal
Because what you make me proof here is the answer to the first question i ask, nope ?

You quoted the new question so, I was not aware of it. Let us see, if:

$\displaystyle s=\min\{k\geq 1:\ker f^k=\ker f^{k+1}\}$

then,

$\displaystyle \ldots \subsetneq \ker f^{s-2}\subsetneq \ker f^{s-1}\subsetneq \ker f^{s}=\ker f^{s+1}=\ker f^{s+2}=\ldots$

and independently on the way you define $\displaystyle r\leq s$ we have $\displaystyle \ker f^s\neq \ker f^r$ if $\displaystyle r<s$ and $\displaystyle \ker f^s=\ker f^r$ if $\displaystyle r=s$ .

10. Ok, now i think i got it.
Maybe it is the way my question is asked which is weird, because it is exactly :

"Proove that, if $\displaystyle r \leq s$, then $\displaystyle \ker f^s = \ker f^r$. Then $\displaystyle s = r$"

So, i've tried something desperate with a lot of complicated things and ended up with... well nothing.
But i think now it is clear.

My excuses for the misunderstanding and my thanking for the help

Hugal.