From: fc03@ns1.cc.lehigh.edu (Frederick W. Chapman)
Newsgroups: sci.math
Subject: Re: polynomial roots and zero divisors
Date: 28 Aug 92 15:25:29 GMT
Organization: Lehigh University

In article <27AUG199211450866@cs.umass.edu>, rcollins@cs.umass.edu
(Bob Collins) writes:

>  I would like to find the eigenvalues/vectors of a square 
>matrix.  Unfortunately, the elements of the matrix are not 
>members of a field.  In particular, they are "dual numbers",
>defined in Yaglom's "Complex Numbers in Geometry" as a type of
>complex number (a + b E), a and b being reals, and E being a
>nilpotent imaginary unit; so E^2 = 0.  Dual numbers form a
>commutative ring, but they are not a field since any number 
>of the form (0 + a E) is a zero divisor.
>
>Question: Can I find the eigenvalues of a matrix (roots of a
>characteristic polynomial) whose elements (coefficients) are 
>from a commutative ring, but not a field?

[...]

Certainly!  In this case, it's not really any more difficult than solving
individual polynomial equations over the reals.  Suppose you want to solve
the following quadratic equation over the "dual numbers":

       (x + y E)^2 + (-3 + E)(x + y E) + (2 + 4 E)  =  (0 + 0 E).

Expand all the multiplications and simplify using the rule E^2 = 0:

         x^2 + 2xy E - 3x + x E - 3y E  + 2 + 4 E  =  0 + 0 E.

Now collect terms together:

            (x^2 - 3x + 2) + (2xy + x - 3y + 4) E  =  0 + 0 E.

Equate the coefficients of the two components to get a system of real
polynomial equations in x and y:

                         x^2 - 3x + 2         =  0
                         (2x - 3)y + (x + 4)  =  0.

The first equation involves only x; solve the first equation to get

                            x = 1  or  x = 2.

Substitute x = 1 into the second equation and solve for y to get y = 5.
Substitute x = 2 into the second equation and solve for y to get y = -6.
Thus, 1 + 5 E and 2 - 6 E are the only solutions to the original quadratic
equation over the dual numbers.

...........................................................................

In general, since 

                (x + y E)^n  =  x^n  +  (n x^{n-1} y) E

and

       (a + b E)(x + y E)^n  =  a x^n  +  (a n x^{n-1} y + b x^n ) E,

we can conclude that any polynomial equation in the unknown x + y E over
the dual numbers will always simplify into a system of two real polynomial
equations in x and y; the first equation will always be a polynomial in x
alone, and the second equation will always be of degree one in y, as
follows:

                        p(x)              =  0
                        p'(x) y  +  q(x)  =  0

where p'(x) denotes the derivative of the polynomial p(x).  Solve the first
equation for REAL solutions x = r, using a suitable numerical method if
necessary.  (To make use of COMPLEX solutions x = u + v I, we must extend
the "REAL dual numbers" to form the "COMPLEX dual numbers"; i.e., numbers
of the form (a + b I) + (c + d I) E, where I^2 = -1, E^2 = 0, IE = EI, and
a, b, c, and d are real; you may or may not wish to make this extension.)

Upon substituting x = r into the second equation, we find that there are
three things that can happen.  (1) If r is NOT a multiple root of p(x) --
that is, if p'(r) is non-zero -- then the second equation has a unique
solution y = -q(r)/p'(r), and r - q(r)/p'(r) E is a solution to the
original polynomial equation over the dual numbers.  (2) If r IS a multiple
root of p(x) -- that is, if p'(r) = 0 -- but r is NOT a root of q(x), then
the second equation has no solutions; thus, there is no real y for which 
r + y E is a solution of the original polynomial equation over the dual
numbers.  (3) If r IS a multiple root of p(x) AND r is a root of q(x), then
all real y satisfy the second equation, thereby yielding an infinite family
of solutions r + y E to the original polynomial equation over the dual
numbers).

Tha tha tha tha tha tha tha tha that's all folks!  :-) 

-- 

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