If $\operatorname{Rank}(A^k) =\operatorname{Rank}(A^{k+1})$, why does $\operatorname{Rank}(A^{k+1}) = \operatorname{Rank}(A^{k+2}) $?

Question

I have found this answer to a very similar question (basically identical).$\DeclareMathOperator{\Rank}{Rank}\DeclareMathOperator{\range}{range}$

However, I can't arrive at the result, even with the hints given.

I can understand the implication is the same as $\range(A^k) = \range(A^{k+1}) \Rightarrow \range(A^{k+1}) = \range(A^{k+2})$.

I can easily show that $\range(A^{k+2})$ is a subspace of $\range(A^{k+1})$, which in turn is a subspace of $\range(A^k)$.

Similarly, I can also see the problem in terms of the kernel of these matrices: $\ker(A^k) = \ker(A^{k+1}) \Rightarrow \ker(A^{k+1}) = \ker(A^{k+2})$

I can easily show that $\ker(A^k)$ is a subspace of $\ker(A^{k+1})$, which in turn is a subspace of $\ker(A^{k+2})$.

I've been trying to solve this for a while, and just can't seem to get there.

I feel like I'm missing something really obvious...

Thanks, Miguel

Answer 1

Here is an elementary proof. Let $x \in \mathrm{range}(A^{k+1})$. Then there is a $y$ such that $x = A^{k+1}y$. Clearly, $A^ky \in \mathrm{range}(A^k)$. Now, since $\mathrm{range}(A^k) = \mathrm{range}(A^{k+1})$, there is a $z$ such that $A^ky = A^{k+1}z$. This implies $$x = AA^ky = AA^{k+1}z = A^{k+2}z,$$ hence $x \in \mathrm{range}(A^{k+2})$. As the other direction is clear, we have $\mathrm{range}(A^{k+1}) = \mathrm{range}(A^{k+2})$.

Answer 2

Hint : As you said, one has $$\lbrace 0 \rbrace \subset \mathrm{Ker}(A) \subset \mathrm{Ker}(A^2) \subset ... \subset \mathrm{Ker}(A^n) \subset ...$$

Try to prove that the sequence $(a_n)_{n \geq 1}$, defined by $a_n = \dim(\mathrm{Ker}(A^n))-\dim(\mathrm{Ker}(A^{n-1}))$, is decreasing.