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