From: israel@math.ubc.ca (Robert Israel)
Subject: Re: General solution of matrix quadratic equations
Date: 26 May 1999 22:54:33 GMT
Newsgroups: sci.math
In article <7ifbrr$dnc$1@pegasus.csx.cam.ac.uk>,
"Adrian Cable" writes:
> Hi there,
> What is known about the general solution of the matrix equation (X^2)a + Xb
> + c = 0, where X is a non-singular (n x n) matrix and a, b, c and 0 are
> column vectors in R^n or C^n? This initially seems like a simple problem,
> but it seems to be fairly resistant to most standard techniques.
As you say, there are a number of cases.
Case 1: If a and b are linearly independent but don't span the whole space,
we can choose Xa to be any vector not in span(a, b), and then the equation
says X(Xa) = -Xb - c, i.e. the result of the transformation X on the vector
Xa is determined in terms of Xb and c. Extend {a, b, Xa} to a basis of your
vector space, and choose the results of X on the other basis vectors
arbitrarily. Thus we have a nonempty (n^2-n)-dimensional family of solutions.
Another family of solutions occurs when Xa is in span(a,b), i.e.
Xa = alpha a + beta b for some scalars alpha and beta. Then
X(Xa) = alpha Xa + beta Xb = -Xb - c, or
(beta+1) Xb = -alpha^2 a - alpha beta b - c. As long as beta <> -1, we
can let alpha and beta be arbitrary, and this determines Xb. Again, the
results of X on the other basis vectors are arbitrary. So this is
a (n^2-2n+2)-dimensional family of solutions.
Case 2: If a and b are linearly dependent but don't span the whole space, say
b = t a (I'll leave the case a = 0 up to you), the first part is similar: Xa can
be chosen to be any vector not in span(a), and X(Xa) = -Xb - c is determined.
We have again a nonempty (n^2-n)-dimensional family of solutions. On the
other hand, if Xa = alpha a for some scalar alpha, then X(Xa) = alpha^2 a
and Xb= alpha t a, so (alpha^2 + alpha t) a = - c. So this family of solutions
may be empty. If it isn't, there are a finite number of possible alpha, and
we have another (n^2-n)-dimensional family of solutions.
Case 3: If a and b are linearly independent and span the whole space (so that
n=2), we must take Xa = alpha a + beta b for some scalars alpha and beta, and
we are as in the second part of Case 1, with a 2-dimensional family of
solutions.
Case 4: If a and b are linearly dependent and span the whole space - well,
that's just n=1.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2