Is it possible to have a $2\times2$ matrix such that $A^3=0\ne A^2$? I know it is possible to have such a matrix if $A$ is $3\times 3$. I think it won't be possible in this case but I'm not sure how to prove a general case.
A specific case proof would work something like this.
$A$ is determined by its action on $e_1,e_2$
Say $Ae_1=\alpha e_1 + \beta e_2$.
Note that $\alpha=0$ else $A^3\ne0$
Similarly $Ae_2=\gamma e_1$
Hence both $\beta=0=\gamma$
How to prove for larger dimension? That is, $A\in M_{n\times n}$ then it is not possible that $A^n\ne0=A^{n+1}$
 A: Consider the images of $A, A^2$ and $A^3$ as subspaces of the plane. If the image of $A$ is the whole plane, then so are the images of $A^2$ and $A^3$. If the image of $A$ is just the origin, then so is the image of $A^2$.
What remains is to study what happens when the image of $A$ is a line $\ell$. Now consider the image of $A^2$. This is equal to the image of $\ell$ under $A$. This is either all the line, or just the origin. If $A\ell$ is just the origin, then that means $A^2=0$ (one application of $A$ sends any point to $\ell$, the second application sends all those points to the origin).
If the image is all of $\ell$, then the image of $A^2$ is the same as the image of $A$, and the image of $A^3$ must again necessarily be the same $\ell$, making $A^3\neq0$.
A: You can handle the general case using Cayley-Hamilton theorem.


*

*Assume there is $A$ with $A^{n+1} =O_{n\times n}$ and $A^n \neq O_{n\times n}$
Using Cayley-Hamilton, this will lead to a contradiction as follows:


*

*According to Cayley-Hamilton we have for the characteristic polynomial $p_A(t) = \sum_{k=0}^na_k t^k$ of $A$:
$$p_A(A) =\sum_{k=0}^na_kA^k = O_{n\times n}$$

*Multiplying this by $A^n$ and using $A^{n+1} = O_{n\times n}$ gives
$$A^np_A(A) = a_0A^n = O_{n\times n} \stackrel{A^n \neq 0}{\Rightarrow} a_0 = 0$$

*The same way, you get successively multiplying by $A^{n-k}, (k=0,\ldots , n-1)$
$$a_1 = \ldots = a_{n-1} = 0$$
It follows
$$p_A(A) = a_n A^n = O_{n\times n} \Rightarrow A^n = O_{n\times n} \mbox{ Contradiction!}$$
A: Start with Cayley-Hamilton for a $2 \times 2$ matrix:
$A^2 - (\text{Tr}A)A + \det A = 0; \tag 1$
multiply it by $A$:
$A^3 - (\text{Tr}A)A^2 + (\det A)A = 0; \tag 2$
use $A^3 = 0$:
$-(\text{Tr}A)A^2 + (\det A)A = 0; \tag 3$
times $A$ again:
$-(\text{Tr}A)A^3 + (\det A)A^2 = 0; \tag 4$
use $A^3 = 0$ again:
$(\det A)A^2 = 0; \tag 5$
since $A^2 \ne 0$, this yields
$\det A = 0; \tag 6$
go back to (3) using this:
$(\text{Tr}A)A^2 = 0; \tag 7$
again, since $A^2 \ne 0$:
$\text{Tr}A = 0; \tag 8$
now use (6) and (8) in (1):
$A^2 = 0 \Rightarrow \Leftarrow A^2 \ne 0; \tag 9$
this contradiction implies there is no such $A$.
Nota Bene: If the Cayley-Hamilton theorem is not available, (1) can be verified by simply grinding out the algebra in terms of the entries of $A$.  Also, I was about to $\LaTeX$ up the general case but trancelocation has it covered. End of Note.
