Endomorphism rank: prove that $r_k-r_{k+1} \geq r_{k+1}-r_{k+2}$

Question

Let $\alpha: V \rightarrow V$ be an endomorphism, and let $r_k = $ rank($\alpha^k$). Show that $r_k-r_{k+1} \geq r_{k+1}-r_{k+2}$.

I have already shown that Im($\alpha^{k+1}$) $\leq$ Im($\alpha^{k}$) and Ker($\alpha^{k+1}$) $\geq$ Ker($\alpha^{k}$) for all $k$, and if $\alpha$ is an automorphism then equality holds throughout; but I can't seem to show why the rank has to obey the above. Rank-nullity doesn't seem to be helping.

Answer 1

Denote $R_k:=\mathrm{im}(\alpha^k)$, and choose a basis for $R_{k+1}$, and extend it by $u_1,\dots,u_s$ to a basis of $R_k$.
Here $s=r_k-r_{k+1}$.

Now apply $\alpha$ to these.
(Observe that $\alpha$ restricted as $R_j\to R_{j+1}$ is surjective, so takes a basis of $R_j$ to a generator set of $R_{j+1}$, for all $j$, use it in particular for $j=k,k+1$.)

Answer 2

If you're comfortable using the Jordan normal form of $\alpha$, then it suffices to note (or show) that $r_k - r_{k+1}$ is equal the number of blocks associated with $\lambda = 0$ that have size at least $k+1$.