Re: Symbolic solution of quadratic matrix equations

From: Robert Israel (israel_at_math.ubc.ca)
Date: 12/01/04 

Date: 1 Dec 2004 01:01:39 GMT

In article <11447b6d.0411301228.3507dbdc@posting.google.com>,
Gregg Kelly <greggk@ozemail.com.au> wrote:
>In answer to Dave's question below, I believe I can shed a little more
>light on this. Rewrite the matrix equation as

>A.X^2 + B.X + C = 0 (1)

>Let k and v be an eigenvalue and eigenvector of X so X.v = k.v
>Then multiplying (1) by v gives

>(A.k^2 + B.k + C).v = 0 (2)

>so that

>det(A.k^2 + B.k + C) = 0 (3)

>This is then a polynomial equation of degree 2.n for the n eigenvalues
>of X. Now we take n solutions k_i of (3) we can then solve (2) for
>v_i.

>Then X is given by the eigen decomposition formula

>X = V.K.V^-1

>where

>K = diagonal matrix k_i
>V = matrix of vectors v_i

... assuming that X is diagonalisable. You may have trouble if
P(k) = det(A.k^2 + B.k + C) happens to have some roots of multiplicity
> 1.

>Which n solutions of (3) should be chosen I don't know. Maybe they all
>give valid solutions, maybe only some of them do ...

If the v_i you choose are linearly independent, this will work.
Note that for an arbitrary choice of n solutions of (3) there is no
guarantee that the vectors v_i are linearly independent. It's easy
to cook up examples where the same vector v_i spans the null space
of k^2 A + k B + C for two different k's. In fact I don't think it's
necessarily true that the null spaces of k^2 A + k B + C for all the
roots of P(k) span C^n. But my guess is that "generically" this all
ought to work.
 
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

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