Does there exist a $2×2$ matrix such that $A^2 \neq 0$ but $A^3=0$? Prove or disprove it.
I figured out that if such matrix exists, then A cannot be invertible, and because A cannot be a zero matrix (easily proven), A must have rank $1$. This can be seen as an extension question of nilpotent matrices. More generally, does there exist a $n × n$ matrix such that $A^n \neq 0$ but $A^{n+1} = 0$?
By Cayley Hamilton theorem, the minimal polynomial divides the characteristic polynomial. The characteristic polynomial of a $n\times n$ matrix is degree $n$. Thus the minimal polynomial cannot be degree $n+1$.
If a $n\times n$-matrix satisfies the equation $A^{n+1}=0$, then it is nilpotent. In this case, we can conclude $A^n=0$, so there is no such matrix.
No, a $n \times n$ matrix which is nilpotent verifies $A^{n}=0$.
To prove this, let $p$ be the least exponent such that $A^{p}=0$ and let $x$ be a $\mathbb{K}^{n}$ vector such that $A^{p-1}x \neq 0$ ($x$ exists, since $A^{p-1}$ is not the zero matrix). You can easily show that the family $(x,Ax,\dotsc,A^{p-1}x)$ of $\mathbb{K}^{n}$ is free, hence $p \leq n$.
If it is matrix over the field - see the previous answers - NO
If it is matrix over the ring - it is possible :)
In general, if $A$ is an $n$ by $n$ matrix, and $A^m = 0$ for some integer $m \geq 1$, then $A^n = 0$. The most intuitive way to see this is by looking at linear transformations instead of matrices. It comes down to a counting argument: you can't count down from $n$ using more than $n$ steps.
Let $V$ be an $n$ dimensional vector space. The choice of a basis of $V$ allows you to identify $n$ by $n$ matrices with linear transformations $V \rightarrow V$. Specifically, if $v_1, ... , v_n$ is your basis, associate to each $n$ by $n$ matrix $A = (a_{ij})$ the unique linear transformation which sends $v_j$ to $a_{1j}v_1 + \cdots + a_{nj}v_n$.
It is enough to prove that if $T: V \rightarrow V$ is a linear transformation, and $T^m = 0$ for some integer $m \geq 1$, then $T^n = 0$.
We may assume from the beginning that $T$ is not the zero transformation, and take $m$ to be the smallest positive integer such that $T^m = 0$. We have an inclusion of subspaces of $V$:
$$\textrm{Im}(T) \supseteq \textrm{Im}(T^2) \supseteq \cdots \supseteq \textrm{Im}(T^{m-1}) \supseteq \textrm{Im}(T^m) = (0)$$
I claim that each of those inclusions is proper. For if, say $\textrm{Im}(T^i) = \textrm{Im}(T^{i+1})$ for some $m > i \geq 1$, then we would have $\textrm{Im}(T^{i+1}) = \textrm{Im}(T^{i+2})$: the right hand side is already included in the left, and conversely if we take an element $T^{i+1}(v)$ in the left hand side, then we can write $T^i(v) = T^{i+1}(w)$ for some $w$, hence $T^{i+1}(v) = T(T^i(v)) = T(T^{i+1}(w)) = T^{i+2}(w)$ is in the right hand side. Iterating this argument, we would have $(0) = \textrm{Im}(T^m) = \textrm{Im}(T^i)$ for $i < m$, contradiction.
Whenever you have a proper inclusion of finite dimensional vector spaces $W_1 \subsetneq W_2$, the dimension of $W_1$ must be strictly less than that of $W_2$. The dimension of $\textrm{Im}(T)$ is at most $n-1$, and since each inclusion above diminishes the dimension by at least one, we are forced to have $\textrm{Im}(T^n) = 0$, i.e. $T^n = 0$.
clearly we have $\mathbb R^2 \subseteq A(\mathbb R^2)\subseteq A^2(\mathbb R^3)$.
If the first inclusion is not proper then $A$ is invertible so $A^3\neq 0$.
Therefore $A(\mathbb R^2)$ has dimension one or zero. It follows that if $A^2$ is not zero then $A(\mathbb R^2)=A^2(\mathbb R^2)= A^3(\mathbb R^2)=\{0\}$
