Given the square matrix $A$ and that $\ker(A^2) = \ker(A^3)$ does this imply that $\ker(A^3) = \ker(A^4)$?

Question

Consider a square matrix $A$ with $\ker(A^2) = \ker(A^3)$. Is $\ker(A^3) = \ker(A^4)$? Justify your answer.

I understand that $\ker(A^3)$ is at least a subset of $\ker(A^4)$. I am just not sure how to show that they are equal given that $\ker(A^2) = \ker(A^3)$ (rather than $\ker(A^2)$ being a subset of $\ker(A^3)$).

I know that if the vector $x$ is in $\ker(A^4)$ then $A^4 x=0$ and $A^2 A^2 x = 0$. From this is is pretty obvious that $\ker(A^2)$ must be a subset of $\ker(A^4)$. I just am not sure where to go from here in proving / disproving what was asked in the question. Any suggestions / hints / advice?

Answer 1

Hint: For any vector $x$ of the appropriate dimension, $x$ is in $\text{ker}(A^4)$ if and only if $Ax$ is in $\text{ker}(A^3)$...

Answer 2

Erick Wong gave a good hint, but since I didn't find the hints particularly illuminating myself, I thought I would add a more explicit answer for future people.

As the OP notes, $\ker A^3 \subset \ker A^4$, since if $A^3x=0$, then certainly $A^4x=A(A^3x)=A0=0$. Thus we just need to show $\ker A^4 \subset \ker A^3$.

Then if $A^4x =0$, then $Ax\in \ker A^3=\ker A^2$. Thus $A^2(Ax)=A^3x=0$. Thus $\ker A^4\subset \ker A^3$, and $\ker A^3 = \ker A^4$.