$a_k = \mathrm{rank}(A^{k+1}) - \mathrm{rank}(A^k)$ is increasing

Question

$a_k = \mathrm{rank}(A^{k+1}) - \mathrm{rank}(A^k)$ is increasing.

This is equivalent to $$\mathrm{rank}(A^{k+1}) + \mathrm{rank}(A^{k-1}) \geq 2 \mathrm{rank}(A^k) $$

which is a consequence of the Sylvester inequality: $$\mathrm{rank}(XZY) + \mathrm{rank}(Z) \geq \mathrm{rank}(XZ) + \mathrm{rank}(ZY)$$ for $ X = Y = A, Z = A^{k-1}$.

Is there a simpler proof of this fact, without using the full power of Sylvester?

Answer 1

Let $V_i$ be the image of $A^i$. Then $a_k = -\dim V_k/V_{k+1}$, so to show $a_k\le a_{k+1}$, it suffices to show that there is a surjective map $V_k/V_{k+1}\to V_{k+1}/V_{k+2}$, and $A$ naturally induces this map.