Show that there is a $k \in \mathbb{N}$, for which applies: $f^{k}(V)=f^{k+l}(V)$ for all $l \in \mathbb{N}.$

Question

I have a problem with the following task:

Let $V$ be finite-dimensional $\mathbb{K}$ -Vector space and $f: V \rightarrow V$ an endomorphism. We define $f^{0}(V)=V$ and $f^{i+1}(V)=f\left(f^{i}(V)\right) .$ Show that there is a $k \in \mathbb{N}$, for which applies: $f^{k}(V)=f^{k+l}(V)$ for all $l \in \mathbb{N}.$

My idea was that it's about concatenations of mappings:

$f^{i}=\underbrace{f \circ \cdots \circ f}_{i \text { times }}$ and $f^{i}(V)$ is the image of $f^{i}$

1) $V=f^{0}(V) \supseteq f^{1}(V)=\operatorname{Image} f$

2) Let $i \geq 1 .$ Statement: Then the following applies $f^{i}(V) \supseteq f^{i+1}(V)$

3) Then we get a descending chain of finite dimensional vector spaces: $$f^{0}(V) \supseteq f^{1}(V) \supseteq f^{2}(V) \supseteq \cdots \supseteq f^{i}(V) \supseteq f^{i+1}(V) \supseteq \cdots$$

At a certain point the remaining components are all the same.

Now I don't know how to go on. Because, I have never mentioned or used k or l anywhere before.

Have someone an idea how to solve it? Thanks for helping me.

Answer 1

You've basically solved the problem already, there isn't much left to do. As you've noticed, the images of the powers of $f$ form a descending chain of subspaces and, because $V$ is finite-dimensional, that chain has to terminate at some point (because $dim(f^i(V))$ can't decrease indefinitely, being a non-negative integer). That means that there exists an integer $k$, such that $f^k(V)=f^{k+1}(V)$ and hence $f^k=f^{k+1}$. This means that however many times you apply $f$ to $f^k$ it will just stay the same, i.e. for any natural $l$, you will have $f^k=f^{k+l}$.

Answer 2

Hint: Since you work with finite-dimensional vector spaces, I'd argue with dimension by the rank-nullity theorem:

$\dim V = \dim (im(f)) + \dim(\ker (f)).$

Thus $\dim V\geq \dim(im(f))$. By iteration, you get a decreasing sequence of dimensions $\geq 0$.