# Induction proof about entries of powers of strictly upper triangular matrix

Let $A$ be a $n \times n$ strictly upper triangular matrix. Prove that, for $k \ge1$, the matrix $A^k$ has the property that $(A^k)_{i,j} = 0$ for all $(i,j)$ with $j-i < k$. Also, show that $A^n = 0$ for any $n \times n$ strictly upper triangular matrix.

I tried to split the sum into two parts for the $k + 1$th term, but now I'm stuck with the induction proof. Anything would be helpful, thanks.

Suppose we want to show that $(A^{k+1})_{i,j}=0$ for all $(i,j)$ with $j-i<k+1$. I will use $(A^k)_i$ to represent the $i$-th row vector of $A^k$, and $A^T_j$ the $j$-th column vector of $A$. Let $j-i<k+1$.
$$(A^k\cdot A)_{i,j}=(A^k)_i \cdot A^T_j=A^k_{i,1}A_{1,j}+A^k_{i,2}A_{2,j}+...+A^k_{i,n}A_{n,j}=A^k_{i,1}A_{1,j}+...+A^k_{i,i+k-1}A_{i+k-1,j}+A^k_{i,i+k}A_{i+k,j}+A^k_{i,i+k+1}A_{i+k+1,j}+...$$
Consider the terms up to $A^k_{i,i+k}A_{i+k,j}$, in the component of $A^k$, the column index is less than or equal to $i+k$, so they are zeros. Now consider the terms after that, in the component of $A$, the column index are $j$'s, and the row index are greater than or equal to $i+k+1$, which give zeros. So the sum is zero.
• You are welcome! Since $(A^n)_{i,j}=0$ for all $j<i+n$, we can set $n$ large enough, for example, the dimension of $A$, so that $j<i+n$ for all $i,j$. Then $A^n$ is a zero matrix. – KittyL Jan 16 '15 at 11:25