Calculating powers of a superdiagonal matrix

**protaktyn** · August 22nd 2011, 12:04 AM

Hello,

while dealing with nonhomogeneous equations with constant coefficients I ecountered a following problem - I need to calculate powers of a given matrix (all powers up to n-1):

$\mathbb N^{n}_{n} \ni \mathbb M_{n}= \begin{bmatrix} 0&n-1&0&0&...&0&0&0&0\\0&0&n-2&0&...&0&0&0&0\\0&0&0&n-3&...&0&0&0&0\\...&...&...&...&...&...&...&...&... \\0&0&0&0&...&0&3&0&0\\0&0&0&0&...&0&0&2&0\\0&0&0& 0&...&0&0&0&1\\0&0&0&0&...&0&0&0&0 \end{bmatrix}$

Is there an easy way to calculate it?

**girdav** · August 22nd 2011, 03:48 AM

We can show by induction on $p$ that $(M_n^p)_{i,j} = 0$ if $i\geq n-p$ or $j\neq i+p$ and $(M_n^p)_{i,i+p}=\prod_{j=0}^{p-1}(n-i-j)$ .

**Opalg** · August 22nd 2011, 03:55 AM

Originally Posted by protaktyn

I need to calculate powers of a given matrix (all powers up to n-1):

$\mathbb N^{n}_{n} \ni \mathbb M_{n}= \begin{bmatrix} 0&n-1&0&0&...&0&0&0&0\\0&0&n-2&0&...&0&0&0&0\\0&0&0&n-3&...&0&0&0&0\\...&...&...&...&...&...&...&...&... \\0&0&0&0&...&0&3&0&0\\0&0&0&0&...&0&0&2&0\\0&0&0& 0&...&0&0&0&1\\0&0&0&0&...&0&0&0&0 \end{bmatrix}$

Is there an easy way to calculate it?

I guess the easiest way is just to do the matrix multiplication to find $\mathbb M_{n}^2$ and $\mathbb M_{n}^3$ . You will then see the pattern for $\mathbb M_{n}^k$ (and if you really want to prove it, you can use induction).

In fact, the nonzero elements of $\mathbb M_{n}^k$ all lie on a single diagonal, namely the diagonal k places above the main diagonal. The elements on this diagonal, reading from top left to bottom right, are

$(n-1)(n-2)\cdots(n-k),\ (n-2)(n-3)\cdots(n-k-1),\ \ldots,\ k!.$

That formula works for $1\leqslant k\leqslant n-1$ . Then $\mathbb M_{n}^k=0$ for $k\geqslant n.$

Edit. Beaten to it by girdav!

**protaktyn** · August 22nd 2011, 04:48 AM

With "i" being the row, or the column index? Just asking because I've seen both ways of indexing...

**girdav** · August 22nd 2011, 04:51 AM

Originally Posted by protaktyn

With "i" being the row, or the column index? Just asking because I've seen both ways of indexing...

"i" is the column index.

**protaktyn** · August 22nd 2011, 06:00 AM

Thank you both.
Somehow though I couldn't get used to the notation of {i,j} being the indexes of the column and row respectively, but I took my time and figured it out.

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