Re: characteristic polynomial

On Feb 1, 9:30 am, "Andreas Thom" <t...@xxxxxxxxxxxxxxxx> wrote:
let p(t) be a monic polynomial with integer coefficients.

i have a question about the richness of the set of matrices that have
p(t) as characteristic polynomial.
first of all, there is a normal matrix A (normal means A*A=AA*) with
characteristic polynomial p(t). (just write the roots on the
diagonal). secondly, there is an integer matrix with characteristic
polynomial p(t). (take the companion matrix).

question 1:
is there a normal matrix with integer coefficients which has
characteristic polynomial p(t).

question 2:
if question 1 fails to be true, does there exist a positive integer k,
such that p(t)^k has the property.

No.

A normal matrix A whose eigenvalues are all on the unit circle is
unitary.
If it is a matrix of integers, each row must have one entry of +/- 1
and
all the rest 0's. If pi(j) is the column index of that entry in row
j,
then pi is a permutation of {1,...,n}. If pi consists of cycles of
lengths
m_1, ..., m_k, then the characteristic polynomial of A is the product
of
t^{m_j} (+/-) 1 for j = 1 ... k. In particular, the eigenvalues of A
are
all the m_j'th roots of 1 or -1 for j = 1 .. k. So if p(t) is a
polynomial,
such as t^2 + t + 1, whose roots are on the unit circle but do not
include
all the m'th roots of 1 or -1 for any m, there is no normal integer
matrix
whose characteristic polynomial is a power of p(t).


Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

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