| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
[from usenet ... definitions on request -- Dan]
Article 7971 of sci.math
Path: ryn.esg.dec.com!shlump.nac.dec.com!decwrl!sun-barr!cs.utexas.edu!swrinde!zaphod.mps.ohio-state.edu!uakari.primate.wisc.edu!aplcen!haven!umbc3!math9.math.umbc.edu!rouben
From: rouben@math9.math.umbc.edu (Rouben Rostamian)
Newsgroups: sci.math
Subject: A linear algebra/geometry problem
Message-ID: <2602@umbc3.UMBC.EDU>
Date: 8 Dec 89 04:32:45 GMT
Sender: newspost@umbc3.UMBC.EDU
Reply-To: rouben@math9.math.umbc.edu (Rouben Rostamian)
Organization: University of Maryland, Baltimore County
Lines: 14
Here is a cute linear algebra/geometry problem that arose today in the course
of my research in wave propagation. I am posting this here as a stimulating
exercise for those interested in such matters.
A real, symmetric, positive-definite matrix A, and a unit vector m are given.
Let E denote the ellipsoid x.Ax=1, and let P denote a plane tangent to E and
perpendicular to m.
Show that the distance of P from the origin equals the square root of the
quadratic form m.Bm, where B is the inverse of A.
Rouben Rostamian Telephone: (301) 455-2458
Department of Mathematics and Statistics e-mail:
University of Maryland Baltimore County rostamian@umbc.bitnet
Baltimore, MD 21228 rostamian@umbc3.umbc.edu
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1158.1 | HPSTEK::XIA | In my beginning is my end. | Sun Dec 10 1989 18:02 | 24 | |
Let t be the distance between the origin and P. Then the vector from
the origin to P (and orthogonal to P) is tm. (1)
Let {y : 1 <= 1 < n} be the set of vectors that are tangent to P.
i
Then it is immediate that tm.y = 0 for all i. Let x be the
i
vector from the origin to E. Now since A is positive definite and
symmetric, A is essentially another metric on the space. Hence, the
Gauss lemma says that radial orthogonality is preserved by the
exponential map. This means x.Ay = 0 for all i. This together with (1)
i
-1
uniquely determines x = (tA) m (note x and tm must be both on P).
By hypothesis and the fact that A and B are symmetric, we have
2 2 2
x.Ax = (Bm).ABm/t = 1 ==> m.BABm/t = 1 ==> m.Bm = t . Q.E.D.
Eugene
| |||||
| 1158.2 | ALLVAX::ROTH | If you plant ice you'll harvest wind | Mon Dec 11 1989 10:15 | 40 | |
A less jargon-y explanation may be appropriate...
If m is any vector lying along the gradient of a quadratic form
at the point x
A x = k m, k is some scalar "scaling" the length of m
The equation of a plane passing thru x a distance d from the origin
and with the same normal vector is
t
m x = d (this is where we assume m is a unit vector.)
t
We can calculate d from k since x A x = 1 and A is symmetric.
t t t t
m = 1/k x A = 1/k x A
t
1/k x A x = 1/k = d
Also since A is positive definite too
-1
x = k A m
t -1
k m A m = d
t -1 2
m A m = d
the desired result.
I suspect that his ellipsoid is an indicatrix of propagation in an
anisotropic medium - that's where this usually comes up (Hamilton's
principle, etc.)
- Jim
| |||||