Some iterate of a linear operator over $\mathbf F_q$ is a projection If $T$ is an endomorphism of a finite-dimensional vector space $V$ over a finite field, then how can I show that there exists a positive integer $r$ such that $T^r$ is a projection operator?
 A: Here is an outline and some hints:


*

*Show that over the finite field $\mathbb F_q$, there exists a number $r(q)$ such that every element of $\mathbb F_q$ is a solution to  $x^{2r(q)}=x^{r(q)}$ (hint: what is the order of $\mathbb F^*_q$)?

*If $A$ is a matrix of a finite field, some power of a $A$ is diagonalizable (hint: it suffices to consider Jordan blocks and use the formula $(a+b)^p=a^p+b^p$ where $p$ is the characteristic).


Note that for step 2, we require the existence of Jordan normal form.  Unfortunately, if the eigenvalues of $T$ don't lie in $\mathbb F_q$, we will have to pass to a larger field which does contain them.  The simplest way is just to adjoin all the eigenvalues to $\mathbb F_q$, which yields a larger finite field.  In particular, this implies that for part 1, we have to replace $q$ with $q^n$ (where $n$ depends on the eigenvalues of $T$).
A: Lemma. Every polynomial $a(t) \in \mathbb{F}_q [t]$ divides $t^{rs} - t^r$, for some $r,s \geq 2$. 
Proof. Pick $n$ such that $\mathbb{F}_{q^n}$ contains every root of $a(t)$ in a fixed algebraic closure $\overline{\mathbb{F}_q}$ of $\mathbb{F}_q$. So $t^{q^n} - t$ contains every prime factor of $a$. Therefore, if $q^m$ is greater than all multiplicities of the roots of $a$, $a \vert (t^{q^n} - t)^{q^m} = t^{q^{n+m}} - t^{q^m}$. Pick $r = q^m$ and $s = q^n$. $\square$
Now, let $f$ an endomorphism of a finite vector space $V$ over $\mathbb{F}_q$. Consider the minimal polynomial $m_f(t)$ of $f$. For the observation, there exist $r, s \geq 2$ such that $m_f(t) \vert t^{rs}- t^r$, therefore $f^{rs} = f^r$. Finally $P = f^{r(s-1)}$ is a projection because $P^2 = f^{rs} f^{r(s-2)} = f^r f^{r(s-2)} = f^{r(s-1)} = P$.
A: You can do this very much in the spirit of Aaron's first hint (but I work with the finite ring of endomorphisms as opposed to the field itself): 
1) The ring of endomorphisms of $V$ is finite.
2) Therefore there are repetitions among the powers $T^r$, $r$ a positive integer.
3) Therefore the sequence $(T^i)_{i\in\mathbf{N}}$ is eventually periodic with a period $\ell$.
4) If $\ell$ is a period, then so are $2\ell,3\ell,\ldots$.
5) $r=k\ell$ works for a large enough value of $k$ such that $T^r$ is in the periodic part of the above sequence of operators.
