Re: Symbolic solution of quadratic matrix equations

From: Dave Rusin (rusin_at_vesuvius.math.niu.edu)
Date: 12/01/04 

Date: 1 Dec 2004 01:04:55 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)

(THat's sort of the transpose of the original equation, but now generalized).

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

Very good! And indeed the same proof shows any set of n roots of your
degree-2n polynomial will work (assuming the roots are distinct). In the
case n=3 this gives C(6,3)=20 possible solutions X, which is what I
computed a different way.

I wrote,

>> The only surprise (to me) is that the polynomial always seems to factor:
>> in the 2x2 case it's the product of a quadratic and a quartic with
>> coefficients in the 7- and 15-digit range, respectively.
>> In the 3x3 case there's a degree-8 factor (60-digit coefficients)
>> and a degree-12 factor (90-digit coefficients). Moreover, these
>> polynomials are "special": the quartic has a dihedral Galois group,
>> and the two factors in the 3x3 case have galois groups of order just 48.

THe factorization seems to be this: when A,B,C are specialized in the
way the OP asked about, Gregg's polynomial is reciprocal (if k is a
root, so is 1/k; that is, it's really a polynomial of degree n in
w = k + 1/k ). It turns out that the octic factor is the one whose
roots correspond to matrices whose eigenvalues involve no reciprocal pairs
(there are 2^3 such choices) and the other factor corresponds to matrices
whose eigenvalues involve one reciprocal pair ( 3 x 4 = 12 such choices ).
I suppose for larger values of n there will be other factors
according to the number of reciprocal pairs of eigenvalues for X.

dave

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