Prove that A^k = 0 implies A^2 = 0 for A_2x2

Question

I shall prove the statement that if A is a 2x2 matrix and $A^2 \neq 0$, then it follows that $A^k \neq 0$ for $k > 2$ as well. I intended to prove the contraposition, but I’m not quite sure how to do it. I know there are already solutions to this question, but I don’t fully understand them. Could anybody help?

Answer 1

How many dimensions has the range of $A$?

If $0$, then $A=0$, and there is nothing to prove.

If $2$, then $A$ is onto, and so $A^k$ is onto for every $k$ by induction.

If $1$, then $A$ has rank $1$, and so the null space has dimension $1$ as well. If the range of $A$ happens to be the null space of $A$, then $A^2=0$. However, if the range of $A$ is not the null space of $A$ then, since both are one-dimensional and thus lines through the origin, their only intersection is $0$. Now fix some $x$ not in the null space of $A$ and observe that $y=A^kx\ne0$ implies $Ay\ne0$ hence $A^{k+1}x\ne0$.