Range of successive linear transformations $T^n$ I've been reading Kato's A Short Introduction to Perturbation Theory for Linear Operators, and I'm stuck on a question in Linear Algebra (specifically on the range of a linear transformation).

Let $T:X\rightarrow X$, $dim(X)$ finite, and $T^n(x)=T(T^{n-1}(x))$. Set $R_n=Range(T^n)$ and $K_n=Ker(T^n)$, $n=1,2,...$ (...) There is a nonnegative integer $m \le dim(X)$ such that: 
  $$R_n \ne R_{n-1}, n<m$$
  $$R_n = R_{n-1}, n\ge m$$

The same exercise showed that $R_n \subseteq R_{n-1}$ and $K_n \supseteq K_{n-1}$, but I'm struggling with the remainder of the question.
 A: Clearly, here (at least in this section, not necessarily in the entire book) vectors spaces are assumed to be finite-dimensional — because otherwise this claim is not true. But for finite-dimensional spaces it's pretty easy. Here's one possible proof.
The numbers $\dim R_n$ are integers that all lie within $0\le\dim R_n\le\dim X$. Being a finite set of integers, it has the smallest element, say $t$. Let $R_m$ be the first (in this chain) subspace with $\dim R_m=t$. Then since $R_{m+1}\subseteq R_m$, we know that $\dim R_{m+1}\le\dim R_m=t$. But $t$ is the least of all these dimensions, so $\dim R_{m+1}=t$. By induction, we have the claim. Similarly for kernels, but using the largest of the dimensions.
But why were all dimensions different before? Assume $R_n=R_{n-1}$ for some $n$. We know that $R_{n+1}\subseteq R_n$. Now pick any $y\in R_n$, then for some $x\in X$ we have $y=T^n(x)=T(T^{n-1}(x))$. Note that $z=T^{n-1}(x)$ lies in $R_{n-1}=R_n$, and therefore $z=T^n(x')$ for some $x'\in X$. Now,
$$y=T^n(x)=T(T^{n-1}(x))=T(z)=T(T^n(x'))=T^{n+1}(x')\in R_{n+1}.$$
Thus $R_n\subseteq R_{n+1}$, which together with the opposite inclusion implies that $R_n=R_{n+1}$. This means that if we have an equality once, them from that point on the sequence of these subspaces has completely stabilized.
This argument, together with the observation that a non-increasing sequence of non-negative integers $\dim R_n$ can't be decreasing forever and thus there has to be at least one place where $R_n=R_{n-1}$, makes my first argument (with the least dimension) unnecessary. But I didn't feel like deleting it... :-)
UPDATE. To address the remaining part: why is $m\le\dim X$? Informally speaking here's the reason. We have the chain of subspaces $X\supseteq R_1\supseteq R_2\supseteq\cdots$ and the corresponding dimensions $\dim X\ge\dim R_1\ge\dim R_2\ge\cdots$. For as long as the dimensions go down, they go down at least by one. But they can't become negative, so going down can only continue for $\dim X$ steps at most.
A more rigorous proof can be done in two cases.
Case 1: $R_1=X$. In this case, $T$ is onto, and for finite-dimensional vector spaces it's equivalent to being bijective, so all $R_n=X$ and $m=1\le\dim X$.
Case 2: $R_1\subsetneqq X$. In this case, the dimensions of all $R_n$ lie within $0$ and $\dim(X)-1$, so we have $\dim X$ possible values. By the pigeonhole principle, the dimensions of the $\dim(X)+1$ subspaces $R_1$, $R_2$, …, $R_{\dim(X)+1}$ can't all be different, so there exists $m$, $1\le m\le\dim X$, such that $\dim R_m=\dim R_{m+1}$ and thus $R_m=R_{m+1}$.
A: If we assume $X$ to be finite-dimensional, the fact that $R_n\supseteq R_{n+1}$ for all $n\in\Bbb N$ implies there must by some integer $n$ where $R_n=R_{n+1}$ (the dimension cannot indefinitely keep decreasing strictly), and one can fix $n_0$ to be the smallest such integer. (In your question you would have $m=n_0+1$; it is unclear to me why one would define $m$ to be one beyond the index where the inclusions cease to be strict.) The point is to prove that also $R_n=R_{n_0}$ for all $n\geq n_0$ (which implies $R_n=R_{n+1}$ for such$~n$).
This is immediate by induction on $n-n_0$ (the starting case $n=n_0$ is obvious). Note that by definition $R_{n+1}=T(R_n)$ (the image under $T$ of the subspace $R_n$) for all $n\in\Bbb N$. Now assuming by induction that $R_n=R_{n_0}$, we get $R_{n+1}=T(R_n)=T(R_{n_0})=R_{n_0+1}=R_{n_0}$, completing the induction step.
