TITLE: How to show that the nth power of a $n \times n$ nilpotent matrix equals to zero $A^n=0$
QUESTION [13 upvotes]: $A$ is a $n\times n$ matrix such that $ A^m = 0 $ for some positive integer $m$.
  Show that $A^n = 0$.

My attempt:
For $n > m$, it's obvious since matrix multiplication is associative.
For $n < m$, $A^n\times A^{m-n} = 0$; not sure what to do next.
Also I know that $\det A = 0$.

REPLY [3 votes]: Since this question is marked as duplicate from one that explicitly asks not to use the Cayley-Hamilton theorem I will not do that. Nor will I use Jordan normal forms (which require an algebraically closed field).
Let $m\in\Bbb N$ be minimal such that $A^m=0$; then by assumption there are (unless $m=0$, which implies $n=0$ and is boring) vectors$~v$ such that $A^{m-1}\cdot v\neq0$; one needs to prove $m\leq n$. There are various ways to proceed, I'll mention three.
(1) Putting $v_i=A^{m-i}\cdot v$ for $0<i\leq m$, one has $v_i\in\ker(A^i)\setminus\ker(A^{i-1})$, which shows that the chain of subspaces $\{0\}=\ker(A^0)\subseteq\ker(A^1)\subseteq\ker(A^2)\subseteq\cdots$ is strictly increasing until reaching $\ker(A^m)=k^n$, and considering dimensions one gets $n\geq m$.
(2) Those vectors $v_1,\ldots,v_m$ are linearly independent, for otherwise, if one considers the first $k$ for which $v_1,\ldots,v_k$ are linearly dependent, then $k>1$ (since $v_1\neq0$ by construction), but applying $A$ to a dependency relation one gets a shorter one, contradicting minimality. From the independence one gets $m=\dim\langle v_1,\ldots,v_m\rangle\leq n$.
(3) Putting $V_i=\ker(A^i)$ for brevity (any $i\in\Bbb N$), each $V_{i+1}$ is (by definition) the inverse image under the linear map defined by $A$ of $V_i$. Then that linear map induces a map $V_{i+2}/V_{i+1}\to V_{i+1}/V_i$ that is injective (by the mentioned inverse image property). It follows that the sequence of natural numbers $(\dim(V_{i+1}/V_i))_{i\in\Bbb N}$ is weakly decreasing, so it cannot attain the value $0$ for any $i<m$. Then immediately $\dim(V_i)\geq i$ for all $i\leq m$, and in particular $n=\dim(V_m)\geq m$. [The main interest of this approach is that it gives additional and quite useful information beyond what was asked for.]