How to prove: if  is a × matrix and $^{}=0$, where  >, then $A^m = 0$ if  is a × matrix and $^{}=0$, where  >, then $A^m = 0$
I haven't covered characteristic polynomials, minimal polynomials, and nilpotent yet.
How to prove it? Could u prove some hint for this question?
 A: For any $k \in \mathbb N$ if $\ker A^k = \ker A^{k+1}$, then $\ker A^l = \ker A^{l+1}$ for $l \ge k$. This can be proven by induction.
$A^n = 0$ means that $\ker A^n$ is equal to the full space $V$. If $r$ is the smallest integer such that $A^r = 0$, we have according to previous result
$$\ker A \subsetneq \dots \subsetneq \ker A^r = V$$
If $r > m$ we would have a sequence of strictly decreasing (for the inclusion) sequence of linear subspaces of length greater than the dimension of the space. A contradiction.
A: Suppose $A^mx \ne 0$, but $A^{m+1}x=0$ for some $x\in\mathbb{R}^m$ (or $x\in \mathbb{C}^m$ if you are working with a complex space.) Then $\{ x,Ax,\cdots,A^mx\}$ is a linearly independent set of vectors. To see why, suppose
$$
                \alpha_0 x + \alpha_1 Ax + \cdots+\alpha_{m}A^{m}x=0.
$$
Apply $A^{m}$ to both sides in order to conclude that $\alpha_0 A^{m}x=0$, which implies that $\alpha_0=0$ because $A^{m}x\ne 0$. Then apply $A^{m-1}$ to both sides to conclude that $\alpha_1=0$. Eventually conclude that
$$\alpha_0=\alpha_1=\cdots=\alpha_m=0.$$
Therefore $\{ x,Ax,\cdots,A^{m}x\}$ is a linearly independent set of vectors, which contradicts the dimension of the vector space that you have.
