Rank sequence - inequality

Question

Consider the sequence $$a_k=\operatorname{rank}(A^{k+1})-\operatorname{rank}(A^k)$$

Prove that the sequence is nondecreasing.

What we want to prove is that $a_{k+1} \geq a_{k}$ or equivalently $$\operatorname{rank}(A^{k+2})+\operatorname{rank}(A^k)\geq2\operatorname{rank}(A^{k+1})$$

I proceeded by induction: for $k=0$ the statement is $$\operatorname{rank}(I_n)+\operatorname{rank}(A^2)\geq2\operatorname{rank}(A)$$ which is just the Sylvester inequality for two identical matrices. Suppose $$\operatorname{rank}(A^{k+2})+\operatorname{rank}(A^k)\geq2\operatorname{rank}(A^{k+1})$$ holds, then we want to prove $$\operatorname{rank}(A^{k+3})+\operatorname{rank}(A^{k+1})\geq2\operatorname{rank}(A^{k+2})$$

That's where I got stuck. I tried approximating $\operatorname{rank}(A^{k+3})$ with Sylvester's inequality as follows:

$$\operatorname{rank}(A^{k+3})\geq \operatorname{rank}(A^{k+1})+\operatorname{rank}(A^2)-n$$

so that

$$\operatorname{rank}(A^{k+3})+\operatorname{rank}(A^{k+1})\geq 2\operatorname{rank}(A^{k+1})+\operatorname{rank}(A^2)-n$$ and then using the base case (for $k=0$) we get $$\operatorname{rank}(A^{k+3})+\operatorname{rank}(A^{k+1})\geq 2\operatorname{rank}(A^{k+1})+\operatorname{rank}(A^2)-n \geq 2\operatorname{rank}(A^{k+1})+2\operatorname{rank}(A)-2n$$ so now it suffices to show that $$\operatorname{rank}(A^{k+1})+\operatorname{rank}(A)-n \geq \operatorname{rank}(A^{k+2})$$ but this can only hold as an equality, since the opposite inequality follows from Sylvester's inequality. In this process, I did not use the inductive hypothesis.

I also arrive to the same conclusion when considering the inequalities $$\operatorname{rank}(A^{k+3})\geq \operatorname{rank}(A^{k+2})+\operatorname{rank}(A)-n$$ $$\operatorname{rank}(A^{k+1})\geq \operatorname{rank}(A^{k})+\operatorname{rank}(A)-n$$ by adding them and using the inductive hypothesis.

How do I proceed from here? Is there a mistake in my train of thought or do I just have to observe something more?

Answer 1

By Rank-nullity theorem it's sufficient (finite-dimensional case) to prove that the sequence $b_k := \dim(B_k) - \dim(B_{k-1})$ are non decreasing.

Let's denote with $B_i=\ker (A^i)$. Since we have $\{0\}=B_0 \subset B_1 \subset B_2 \subset \dots \subset B_n \subset \cdots $ we have the following : $\forall 1 \le i \le n-1, \dim(B_i) - \dim(B_{i-1}) \geq \dim(B_{i+1}) - \dim(B_i)$, i.e $b_i \geq b_{i-1}$.

Consider the following homomorphism :

$$\begin{array}{ccccc} & B_{i+1} & \xrightarrow{f} & B_i & \xrightarrow{\pi_i} & B_i /B_{i-1} \end{array}$$

$\ker(\pi_i \circ f)=f^{-1}(\pi_i^{-1}(\{0\}))=f^{-1}(B_{i-1})=B_i$, so by 1-st isomorphism theorem we have $B_{i+1} /B_{i} \simeq \text{Im}(\pi_i \circ f) < B_i /B_{i-1}$. So $\dim(B_{i+1} /B_{i}) \le \dim(B_i /B_{i-1})$ which was what we wanted to prove.

Answer 2

I would proceed as follows.

As $\operatorname{Im}(A^{k+1}) \subseteq \operatorname{Im}(A^{k}) $, one can find a linear subspace $G$ of $\operatorname{Im}(A^{k}) $ such that $\operatorname{Im}(A^{k}) =\operatorname{Im}(A^{k+1}) \oplus G $. Applying $A$ on both sides of the last equality $$A(\operatorname{Im}(A^{k}))=\operatorname{Im}(A^{k+1})=A(\operatorname{Im}(A^{k+1})) +A(G) = \operatorname{Im}(A^{k+2}) +A(G)$$ and taking the dimensions, we get $$\begin{aligned}\operatorname{rank}(A^{k+1})-\operatorname{rank}(A^{k+2}) &= \dim A(G) - \dim \operatorname{Im}(A^{k+2}) \cap A(G)\\ &\le \dim A(G)\\ &\le \dim G = \operatorname{rank}(A^{k})-\operatorname{rank}(A^{k+1}) \end{aligned}$$ as desired.

Answer 3

From Frobenius inequality we have $$\operatorname{rank}(XYZ)+\operatorname{rank}(Y) \geq \operatorname{rank}(XY)+\operatorname{rank}(YZ)$$

Set $X=Z=A, Y=A^k$ so that $XYZ=A^{k+2},XY=YZ=A^{k+1}$. But then:

$$\operatorname{rank}(A^{k+2})+\operatorname{rank}(A^k) \geq 2\operatorname{rank}(A^{k+1})$$ for all $k$ and thus we are done.

Ah, it was so obvious! Thank you everyone for your great answers!