TITLE: Range of successive linear transformations $T^n$
QUESTION [1 upvotes]: 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.

REPLY [1 votes]: 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.
{"set_name": "stack_exchange", "score": 1, "question_id": 2223189}