TITLE: How to prove a matrix is nilpotent?
QUESTION [17 upvotes]: 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?

REPLY [5 votes]: 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 $A^n = 0$.