Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So there is a formula for the $n$th power of a matrix in Jordan normal form. Is there a formula for the $n$th power of a general triangular matrix? If not, are there known formulas for "nice" upper triangular matrices? Like those consisting of all 1s and 0s.

share|improve this question
1  
For an upper triangular matrix that can be written as $D + N$ where $D$ is diagonal and $N$ is strictly upper-triangular (hence nilpotent) and such that $D$ and $N$ commute (in particular if $D$ is scalar), it is very easy to compute the powers explicitly in terms of the powers of $D$ and $N$. In general I think you should just find the Jordan normal form or use special structure of the matrix in question. –  Qiaochu Yuan Oct 5 '11 at 1:07
    
Well, the diagonal elements are easy, and there's a neat formula for the superdiagonal entries, involving sums of the form $\sum\limits_{k=0}^{n-1} x^k y^{n-k-1}$... –  J. M. Oct 5 '11 at 1:09

1 Answer 1

If $A$ is an $n \times n$ upper triangular matrix, $m$ is a positive integer and $i \le j$, $(A^m)_{ij} = \sum \prod_{k=1}^m A_{i_{k-1} i_k}$ where the sum is over all nondecreasing $m+1$-tuples
$i = i_0 \le i_1 \le \ldots \le i_m = j$ starting at $i$ and ending at $j$. In particular, if $A$ consists only of 1's and 0's, $(A^m)_{ij}$ is the number of $m$-step paths from $i$ to $j$ in the directed graph with adjacency matrix $A$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.