Suppose $A$ is a $2\times 2$ matrix with $A^k = 0$. Then $A^2 = 0$. $$A =   \left(\begin{matrix}
        a & b \\
        c & d \\
        \end{matrix}\right)
$$
a) Prove that we have  $ A^2-(a+d)A + (ad-bc)I_2 = O_2.$
b)  Show that if there exists an integer $k\ge2$ such that $A^k = O_2$ then $A^2 = O_2.$
Part a) was very easy to show (just simple matrix operations). Part b) is what I'm struggling with...
 A: There are very many ways to do (a). If you are familiar with eigenvalues and Cayley-Hamilton, then you can recognize this as exactly the statement of the Cayley-Hamilton theorem in for 2x2 matrices. Supposing that you are not familiar with Cayley-Hamilton, then you can just do it out by hand. It's not particularly hard, it's just not particularly interesting.
There are also very many ways to do (b). Nimda gave one proof from a slightly high point of view in the comments. But it can also be done from first principles.
We'll consider the rank of the matrix $A$. If $A$ has full rank, then it cannot be that $A^k = 0$. If $A$ has zero rank, then it is trivial that $A^k = A^2 = A = 0$. So we only need to consider the case when $A$ has rank one.
So there is a vector $x$ so that $Ax \neq 0$. As $A^k x = 0$, at least one of $\{ Ax, A^2x, \ldots, A^{k-1}x\}$ is a nonzero vector, which I'll denote by $v$, that $A$ maps to $0$. Then $v$ is a nonzero vector in the image of $A$ and also in the kernel of $A$. Since the rank of $A$ is one, the dimension of the image and kernel are the same, and so we conclude that the image of $A$ is the same as the kernel of $A$.
And thus $A^2 x = 0$ for all $x \in \mathbb{R}^2$. $\diamondsuit$
A: If $A^k = 0$ it means that $|A|^k = 0$ and so $|A|=0$. Thus, you can say $ad - bc = 0$. If $k=2$ we are done. Otherwise, from part (a), you get $A^2 = (a+d)A$. If $(a+d) = 0$, again we are done! Assume $a+d \neq 0$. Multiply both sides by $A^{k-2}$ to get $A^k = (a+d)A^{k-1}$. Therefore, $A^{k-1} = 0$. You can continue this procedure $k-2$ times to get the result!
