Re: Stability of a Companion Matrix

Correction:
z^n-Sum[x[m+1]*z^m,{m,0,n-1}]=0
A simple solution that gives a Salem is:
x_1=1
x_n=2
x_n-1=...x_2=0
Mathematica:
M = {{0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0},
{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1},
{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2}}
Det[M - x*IdentityMatrix[12]]
-1 - 2* x^11 + x^12
NSolve[Det[M - x*IdentityMatrix[12]] == 0, x]
All but the root 2.00049 are inside the unit disk.
You might be more interested in the oddversion of these
matrices.
Roger Bagula wrote:
Erasmus wrote:

Linearizing a nonlinear difference
equation within neighborhood of its fixed
point yields an n x n companion matrix A =

[ 0 1 0 ... 0 ]
[ 0 0 1 ... 0 ]
[ 0 0 0 ... 0 ]
[ ... ... ... ... ... ]
[ 0 0 0 0 1 ]
[ x_1 x_2 x_3 ... x_n ]

where each of the elements of the last row is

x_i = [1 - 1 / (1 + r)] a_i
a_i = any real number

for i = 1 ... n.

If r = 0, then x_i = 0 for all i and
the matrix has all eigenvalues at the
origin.

My question is, what is the
range of r such that every eigenvalue
of the matrix is within the unit circle?

Thanks.


The characteristic polynomial :
z^n-Sum[x[m+1]*z^m,{m,0,n}]=0
You have stated more or less the Salem Polynomial problem
for the x[n] being Integers.
You want all the roots inside the unit disk right?
It might be easier for real number x[n],
than integers... but I would look at Salem numbers if I were you.
.