Prove that $A^k = 0 $ iff $A^2 = 0$ Let $A$ be a $  2 \times  2 $ matrix and  a positive integer $k \geq 2$. Prove that $A^k = 0 $ iff $A^2 = 0$.
I can make it to do this exercise if I have $ \det (A^k) = (\det A)^k $. But this question comes before this.
Thank you very much for your help!

The standard and more general approach to solve this kind of problems is to consider the minimal polynomial of $A$ (see the other answers here). However, in this particular problem, as $A$ is only $2\times2$, one may solve the problem as follows. As the forward implication is trivial, we consider only the backward implication
Proof 1. If $A^k=0$ for some $k\ge2$, then $A$ is singular and it has at most rank 1. Therefore, $A=uv^T$ for some vectors $u$ and $v$. Note that for any $n\ge1$,
$$
A^n=\underbrace{(uv^T)(uv^T)\cdots(uv^T)}_{n \text{ times}}
=u\underbrace{(v^Tu)}_{\text{scalar}}\cdots(v^Tu)v^T=(v^Tu)^{n-1}A.
$$
In particular, $A^k=(v^Tu)^{k-1}A$ and $A^2=(v^Tu)A$. Now the result follows immediately.
Proof 2. If $A^k=0$ for some $k\ge2$, then $A$ is singular.
Let $Ax=0$ for some $x\ne0$. Extend $x$ to a basis $\{x,y\}$ of the underlying vector space. Let $Ay=px+qy$. By mathematical induction, we have $A^ny=pq^{n-1}x + q^ny$ for any $n\ge1$. Now use $A^k=0$ to argue that $q=0$. Hence $A^2x=A^2y=0$.
