The matrix $A = (a_{i,j})$ in which $a_{i,j} = 0$ if $i ≥ j$ is nilpotent. A square matrix $A$ is called nilpotent if $A^m = 0$ for some positive integer $m$. Show that
A $n \times n$ matrix $A = (a_{i,j})$ in which $a_{i,j} = 0$ if $i ≥ j$ is nilpotent.
 A: A direct argument may be obtained by arguing by induction. You should try the small cases $n = 2, 3, 4$ to see what this means in practice.
Fix $k$. Say that a matrix $B$ satisfies condition null-$k$ if $b_{ij} = 0$ if $i \ge j - k$. (Clearly $B = A$ satisfies null-$0$.)
Suppose $B$ satisfies null-$k$.
Consider the product $C = A B$, and let $c_{ij}$ be one of its elements, with $i \ge j - k - 1$. Then
$$
c_{ij} = \sum_{s=1}^{n} a_{is} b_{sj}.
$$
If $s \le i$, then $a_{is} = 0$. If $s > i$, then $s > i \ge j - k - 1$, that is, $s \ge j - k$, so that $b_{sj} = 0$.
In any case, $c_{ij} = 0$ for $i \ge j - (k + 1)$, that is, $C = A B$ satisfies null-$(k+1)$.
Now $A$ satisfies null-$0$, so that $A^{2}$ satisfies null-$1$, etc, until $A^{n}$ satisfies null-$(n-1)$, which means $A^{n} = 0$.
A: This is obvious  using  Hamilton-Cayley's theorem: the  only eigenvalue of such a matrix is $0$, since its characteristic polynomial is $\chi_A(x)=(-1)^nx^n$. By Hamilton-Cayley, $$\chi_A(A)=0=(-1)^nA^n.$$
