Re: Stability of a Companion Matrix

James, thanks for your reply.

a_i is arbitrary, but r can be a function of a_i, max or min
of them, sum of them, etc. It seems that there must be a
range of r depending on all a_i that will make all the
eigenvalues of the matrix contained in the unit circle.

"James Meiss" <jdm@xxxxxxxxxxxxxx> wrote in message
news:jdm-EFD066.10591112032006@xxxxxxxxxxxxxxxxxxxxxxx
In article <WyIQf.9951$CI6.73@trnddc07>, "Erasmus"
<a@xxxxx> 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.

It seems to me that since your a_i are arbitrary, then the
characteristic polynomial for your A is arbitrary. So all
things are
possible and the value of r is irrelevant.

But maybe I'm missing something.

--
James Meiss
<http://amath.colorado.edu/faculty/jdm>


.

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