If $a_k = \operatorname{rank}(S^k)$, deduce that $a_k - a_{k+1} \geq a_{k+1} - a_{k+2}$ 
Let $S: V \to V$ be a linear transformation. Show that $V \supseteq \operatorname{Im}(S) \supseteq \operatorname{Im}(S^2) \supseteq \cdots$ and $\{0\} \subseteq \ker(S) \subseteq \ker(S^2) \subseteq \cdots$.
If $a_k = \operatorname{rank}(S^k)$, deduce that $a_k \geq a_{k+1}$ and that $a_k - a_{k+1} \geq a_{k+1} - a_{k+2}$.

I can show everything except for $a_k - a_{k+1} \geq a_{k+1} - a_{k+2}$, which eludes me. I attempted using the rank-nullity theorem, but no luck so far. Any ideas/hints?
 A: Hint: Show that the natural projection map $\text{Im}(S^k)/\text{Im}(S^{k+1}) \longrightarrow \text{Im}(S^{k+1})/\text{Im}(S^{k+2})$ is surjective!
A: I prefer to look at the sequence of kernels, as that is what one uses more often (for instance when putting some internal structure in generalised eigenspaces). Putting $n=\dim V$ et $b_k=n-a_k$ one has $b_k=\dim(\ker(S^k))$ by rank nullity, and the sequence $(b_k)_{k\in\Bbb N}$ will be concave rather than convex: $b_{k+1}-b_k\geq b_{k+2}-b_{k+1}$. To show it, note that $S$ restricts to a map $\ker(S^{k+2})\to\ker(S^{k+1})$, which can be composed with the canonical projection $\ker(S^{k+1})\to \ker(S^{k+1})/\ker(S^k)$. The kernel of the composite map $\ker(S^{k+2})\to\ker(S^{k+1})/\ker(S^k)$ is $\{\, v\in V\mid S(v)\in\ker(S^k)\,\}=\ker(S^{k+1})$, so that composite map factors as the canonical projection canonical projection $\ker(S^{k+2})\to \ker(S^{k+2})/\ker(S^{k+1})$ followed by an injective map (said to be induced by $S$) $\ker(S^{k+2})/\ker(S^{k+1})\to \ker(S^{k+1})/\ker(S^k)$. Taking dimensions of the spaces, that injectivity means $b_{k+2}-b_{k+1}\leq b_{k+1}-b_k$, which is what we were wanted to prove.
