Date: Thu, 6 Apr 95 03:03:20 CDT
From: rusin (Dave Rusin)
To: jnygaard@math.uio.no
Subject: Re: Chebychev companion matrix
>Thanks for your answer. What you describe works well with the
>coefficients of the polynomial p(x)=a_0+a_1x+...+a_nx^n, but the point
>is that my polynomial is given as p(x)=c_0+c_1T_1(x)+...+c_nT_n(x),
>T_i(x) Chebychev polynomials, and I don't want to compute the a's from
>the c's. Obviously, (it seems to me) doing what you prescribes with
>the c-coefficients, does not give the roots of p(x). (What it gives, I
>cannot immediately see.)
Right. I don't rememebr just what the T_i are, but perhaps this will help.
If p is any polynomial, then the quotient ring V = C[x]/(p(x)) is a
finite dimensional vector space; indeed, if p has degree n, then V
is spanned by 1, x, x^2, ..., x^(n-1). The way the companion matrix comes
into play is that it's the matrix that goes with the linear transformation
which multiplies everything by x. (If p = Prod(x-r^i), then another
basis is the set of polynomials bi(x)=p(x)/(x-ri), which have the property
that x.bi = ri.bi mod p(x), so that the bi are the eigenvectors
and the eigenvalues are precisely the ri. That's why the companion matrix
as I just described it has the right eigenvalues.)
So perhaps what you want to do is to find a nice way to describe
x*T_i in terms of the other T_j ? Then you would have yet another
representation of the same linear transformation in another basis, so that
the new matrix again has the right minimal polynomial.
I'm sorry, but I cannot recall the recursion for the T_i (there's one
of each degree, right? Determined by some orthogonality relation?)
dave