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- From: sec@otter.hpl.hp.com (Simon Crouch)
- Date: Fri, 28 Aug 1992 07:13:11 GMT
- Subject: Re: polynomial roots and zero divisors
- Message-ID: <640033@otter.hpl.hp.com>
- Organization: Hewlett-Packard Laboratories, Bristol, UK.
- Path: sparky!uunet!usc!sdd.hp.com!hpscdc!hplextra!otter.hpl.hp.com!otter!sec
- Newsgroups: sci.math
- References: <27AUG199211450866@cs.umass.edu>
- Lines: 27
-
- In sci.math, 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.
-
- Bob,
- I'm unfamiliar with Yaglom's definition, but you seem to be using
- a very simple case of "superspace" (the definition of it that
- is the algebra generated by 1 and anticommuting b_{1}....b_{n})
- with one anticommuting generator. There are various sources
- that go through linear algebra in superspace including
- "superdeterminants" and "superinverses". DeWitt's book "Supermanifolds"
- and a paper by, I think, Leites in the Doklady about 15 years ago
- cover this stuff (sorry for being vague but my references are at home!).
- I had some fun in my doctoral thesis analysing "supermanifolds" based
- on numbers of the form (a + b E), E^2 = 0.
-
- Simon.
-
-
-