Suppose we have the following matrix with real entries $k_i \geq 0$: $$ K=\begin{pmatrix} 0 & & \ldots & 0 & k_1 \\ k_2 & 0 & \ldots & 0 & 0 \\ 0 & k_3 &\ldots & 0 &0\\ \vdots & \vdots & \ddots & 0& \vdots \\ 0 & 0 & \ldots & k_N & 0 \end{pmatrix} $$ I want to show that $K^N = \kappa^N I$, where $\kappa = \left(\prod\limits_{i=1}^{N} k_i\right)^{(1/N)}$ is the geometric mean of the entries. In other words, the $N$-th power of the matrix $\frac{1}{\kappa}K$ is the identity matrix.

Using a few lines of code I can compute these matrices and check that this is true for any $N$ I have tried. Now I would like to prove it without the need of numerical computation.

What I've tried is to write the matrix entries as $K_{ij} = k_i (\delta_{i,j+1} + \delta_{i,j-N+1})$, do matrix multiplication with the Kronecker deltas and discard the entries that do not satisfy $1 \leq i,j \leq N$. However, after just a few multiplications the terms become too cumbersome to work with.

One property of the above matrix is that it is a generalized permutation matrix, but I do not know of any properties from such matrices that could be used to prove my statement.

I am also aware of this question on invertibility of a square matrix with integer entries. However, I do not only want to just prove invertibility but show that a particular power of a matrix is the identity.


Perhaps the simplest thing is to write the action of $K$ on the standard basis $e_{i}$.

This is just $K e_{i} = k_{i+1} e_{i+1}$, with the convention that $N + 1$ is replaced by $1$, and in general $N+j$ by $j$. In other words, indices are taken modulo $N$.

Then for all $i$ $$ K^{N} e_{i} = K^{n-1} k_{i+1} e_{i+1} = K^{n-2} k_{i+1} k_{i+2} e_{i+2} = \dots = \left( \prod_{i=1}^{N} k_{i} \right) e_{i}. $$

A formal proof can be done by induction, with the above convention on the indices: one proves easily that $$K^{t} e_{i} = \left(\prod_{l=1}^{t} k_{i+l}\right) e_{i+t}.$$


I'd rather not use a sum. It obscures which products of Kronecker symbols are null. I'll just do it by cases. The element of $K$ are :

\begin{gather} K_{i,j} = \begin{cases} k_i \text{ if } i\equiv j+1 \mod n,\\ 0 \text{ else.} \end{cases} \end{gather}

Thus the first product gives :

\begin{align} (K^2)_{i,j} &= \sum_{l} K_{i,l}K_{l,j} = \begin{cases} K_{i,i+1}K_{i+1,i+2} \text{ if } i \equiv j+2 \mod n, \\ 0 \text{ else.} \end{cases}\\ &= \begin{cases} k_i \cdot k_{i+1} \text{ if } i \equiv j+2 \mod n, \\ 0 \text{ else.} \end{cases} \end{align}

And so on, by recursion you can deduce that (I won't write the module in the indices any more)

\begin{align} (K^p)_{i,j} = \begin{cases} \prod_{l=i}^{i+p-1} k_l \text{ if } i = j+p, \\ 0 \text{ else.} \end{cases} \end{align}

Of course working modulo $N$, in the case that $p = N$, the condition $i=j+N$ is equivalent to $i = j$. Also, all products all become \begin{gather} \prod_{l=1}^{N} k_l \end{gather}


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