Endomorphisms and projection operators in vector spaces over finite fields In the process of studying for a qual I've become hung up on the following simple problem:
Let $T$ be an endomorphism of a finite-dimensional vector space over a finite field. Show that there is a projection operator $P$ (with $P^{2}=P$) and an integer $r$ such that $P=T^{r}$.
I've worked out the following: Let ${A}\in M_n$ be a matrix representation of $T$ with respect to some basis, where $n$ is the dimension of the vector space $V$ for which $T$ is an endomorphism. Since the field over which $V$ lies is finite, the set $\{A^{k} | k\in\mathbb{N}\}$ is finite. Hence, there exist integers $r,s\in\mathbb{N}$, $r<s$, such that $A^{r}=A^{s}$. Then $A^{s}=A^{s-r}A^{r}=A^{s-r}A^{s}=A^{k(s-r)}A^{s}$ for any $k\in\mathbb{N}$.
And I've played every which way with that but can't seem to find the trick.

I don't have enough reputation to post this as a comment, but I think you are almost done.
Namely, you just showed that, if you start with $A$ and start taking powers, you will get
i) An initial segment $A^{0},\ldots,A^{s}$ for some $s$.
ii) Then, you'll have a segment $A^{s+1},\ldots,A^{s+k}$ where $A^{s+k}=A^{s}$. From here on out, you are stuck in a loop.
I think the last small detail is that the loop part gives you more than you are using. Namely, we know that $A^{s'}=A^{s'+k}$ for all $s'\geq s$. Then, if you pick $s'\equiv 0\pmod{k}$ and $s'\geq s$, we find that $A^{2s'}=A^{s'}$, which seems to be what you wanted.
