From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: homework help
Date: 21 Aug 2000 15:23:26 -0400
Newsgroups: alt.math.undergrad,sci.math
Summary: [missing]

In article <39A1732F.162B1544@home.com>,
Jon and Mary Frances Miller  <jonathanemiller@home.com> wrote:
:I'm helping my daughter with her homework, and I'm stuck on this.
:
:Suppose A is a real skew-symmetric matrix.  Show that both I-A and I+A
:are nonsingular and that (I-A)*(I+A)^-1 is unitary (multiply it by its
:transpose and you get I).
:
:I can show for dim A = 2 and 3 by computing the inverses and the
:transposes and multiplying, but get no hints for how to prove for
:general dim A = n.

The old problem: it is easy to help if the helper is told how things were
defined, and how much has been covered prior to the problem at hand.

Guessing that the Euclidean norm is a known concept, in the form

||x||^2 = x'*x   (prime denoting transpose)

and that the "socks and boots" rule for the transpose of the product is
known:  (P*Q)' = Q' * P' (plus associative law etc.)

Matrix A being skew-symmetric means A'=-A.

Let me start: Show that I+A is invertible by showing that it is
one-to-one. So, for a non-zero vector x,

||(I+A)*x||^2 = ((I+A)*x)' * (I+A)*x = x' * (I+A)' * (I+A) * x

= x' * (I-A) * (I+A) * x = x' * (I-A^2) * x = x' * (I + A'*A) * x

= x'*x + x'*A'*A*x = ||x||^2 + ||A*x||^2 >= ||x||^2

Summarizing:  ||(I+A)*x|| >= ||x|| > 0 since x was not 0.

We found: if x is not 0 then (I+A)*x is not zero, either, so (I+A) is a
one-to-one square matrix, hence invertible.

Then you realize that (I-A) behaves the same way since (-A) is also
skew-symmetric.

Next you will need the fact that (I+A) commutes with (I-A), and that the
transpose of the inverse is the inverse of the transpose (if it exists),
and the rest is mechanical algebra.

Good luck, ZVK(Slavek).