How to prove a matrix is nilpotent? 
Given an $n\times n$ upper triangular matrix $A$ with zero on main diagonal, show that $A^n = 0$.

I did some matrix operation and noticed that the diagonal moves up, ultimately all entries will be zero. Is there a nicer way to do it?
 A: Suppose that $a_{i\,j}=(A)_{i\,j}=0$ for $i\ge j$ and $b_{i\,j}=(B)_{i\,j}=0$ for $i\ge j-m$, where $A$ and $B$ are $n\times n$ matrices. Consider when it is possible to have $a_{i\,j}b_{j\,k}\not=0$.
To have $a_{i\,j}\not=0$ we must have $i<j$ (i.e. $i\le j-1$).
To have $b_{j\,k}\not=0$ we must have $j<k-m$ (i.e. $j\le k-m-1$).
Therefore, to have $a_{i\,j}b_{j\,k}\not=0$, we must have $i<k-m-1$ (i.e. $i\le k-m-2$).
Thus, for $i\ge k-m-1$
$$
(AB)_{i\,k}=\sum_{j=1}^na_{i\,j}b_{j\,k}=0\tag{1}
$$
The matrix $A$ specified in the question has $a_{i\,j}=0$ for $i\ge j$. Using $(1)$ and induction, we have that $(A^m)_{i\,j}=0$ for $i\ge j-m+1$. This means that $(A^n)_{i\,j}=0$ for $i\ge j-n+1$, which means that $(A^n)_{i\,j}=0$ for all $1\le i,j\le n$.
Therefore, $A^n=0$.
A: Here's a nice graph theoretical proof. Given a $n \times n$ matrix $A$ consider the directed graph $D(A)$ on $n$ vertices where vertex $i$ is joined with vertex $j$ with an edge of weight $a_{ij}$ whenever $a_{ij} \neq 0$. 
Define the weight of a walk in $D(A)$ as the product of all the entries on its edges. It can be easily seen that the $ij$ entry of $A^k$ is the sum of weights of all possible walks of length $k$ from vertex $i$ to vertex $j$ in $D(A)$. 
Now if $A$ is a strictly upper triangular matrix then in $D(A)$ there is an edge between vertex $i$ and vertex $j$ only if $i < j$. 
Therefore $D(A)$ does not contain any cycles. And hence there cannot be any walks of length greater than or equal to $n$ in $D(A)$, proving that $A^n = 0$. 
A: Yes, if you already covered this topic....
Hint: What are the eigenvalues of your matrix?
A: If $\mathbf{e}_1,\ldots,\mathbf{e}_n$ is the standard basis for $\mathbb{R}^n$ (or whatever your base field is), then notice that $A\mathbf{e}_1=\mathbf{0}$, and
$$A\mathbf{e}_i \in \mathrm{span}(\mathbf{e}_1,\ldots,\mathbf{e}_{i-1}),\quad i=2,3,\ldots,n.$$
Now show that $A^n\mathbf{e}_i = \mathbf{0}$ for all $i$ to get the desired conclusion.
A: It can be showed by induction: for $n=2$ it's a simple computation. If the result is true for $n$, and $A=\pmatrix{T_n&v\\\ 0&0}$ where $T_n$ is a $n\times n$ triangular matrix with zeros on the main diagonal, then for each $p\geq 1$ we have $A^p=\pmatrix{T_n^p&T_n^{p-1}v\\\ 0&0}$. Therefore, for $p=n+1$ we get $A^{n+1}=\pmatrix{T_n^{n+1}&T_n^nv\\\ 0&0}=0$ (because $T_n^n=0$).
A: An $n\times n$ upper triangular matrix $A$ has characteristic polynomial $X^n$. Thus by the theorem of Cayley-Hamilton you get $A^n=0$.
