I am having difficulty with a proof for an Algebraic Topology
independent study course.
Theorem: Given an nxn matrix, A, with positive, real entries, use
Brower's Fixed Point Theorem to show that A has at least one positive,
real eigenvalue.
My first instinct was to consider A as a linear transformation from Rn
to Rn, then Brower guarentees you at least one fixed point, but the
origin is fixed, so no help there. I can use the trace of the matrix
to show the existance of a positive eigenvalue, but I can not see how
to show a real eigenvalue, or use Brower at all.
Please Help. 
Request for Question Clarification by
mathtalkga
on
20 May 2004 11:05 PDT
Rather than apply Brouwer's Fixed Point Theorem to the ndisk here:
http://en.wikipedia.org/wiki/Brouwer_fixed_point_theorem
you should consider the (n1)disk formed by the intersection of the
positive orthant in R^n and the boundary of ndisk.
Show that the mapping induced by centrally projecting Ax onto this
(topological) (n1)disk for every x in that disk satisfies the
conditions of Brouwer's theorem.
In particular you need to argue that Ax is bounded away from 0 and
remains in the positive orthant, so that projection onto the boundary
of the sphere is welldefined and continuous.
Let me know if you'd like more detailed discussion.
regards, mathtalkga

Clarification of Question by
topologystudentga
on
20 May 2004 12:16 PDT
Your answer really helped. To make sure that I understand,
We have a central projection P onto the (n1) disk D and the induced
function F:D to D given by F = P(Ax) for all x in D.
Using Brower in the positive Orthant, we have a positive, real fixed
point t, and therefore At = t = 1*t and therefore A has an eigenvalue
of 1.
Correct?

Request for Question Clarification by
mathtalkga
on
20 May 2004 13:24 PDT
Well, we cannot say what the positive eigenvalue of A is from this
analysis, and certainly cannot know that it is 1.
By central projection onto the (n1)disk is merely meant to normalize, ie.
F(x) = Ax/Ax
to a unit vector. Now a fixed point of F will satisfy:
x = Ax/Ax
but Ax = Ax x leaves wide open what (positive real) eigenvalue Ax might be.
regards, mathtalkga

Request for Question Clarification by
mathtalkga
on
20 May 2004 20:10 PDT
Perhaps I need to make the (n1)disk a little clearer. I was
thinking about the portion of the "unit" ndisk B_n that lies inside
the positive orthant:
{ (x1,x2,...,xn)  xi >= 0 for each i = 1,..,n }
Since you are studying algebraic topology, it might be more convenient
to use instead the (n1)simplex in standard position, ie. the convex
hull of the unit vectors in R^n. Topologically this is an (n1)disk,
and every nonzero point in the positive orthant has a "unique
projection" onto this simplex from scaling by the sum of coordinates:
(x1,x2,...,xn) > (x1,x2,...,xn)/(SUM xi)
Of course this is essentially the same as scaling by the Euclidean
norm to map nonzero vectors in this "cone" onto the unit sphere in
R^n, as I was outlining before.
The important thing is to recognize that the resulting composition of
maps (first multiplying by A, then projecting back to the disk) is
continuous on the disk. Then the hypotheses of Brouwer's fixed point
theorem are fulfilled.
regards, mathtalkga

Clarification of Question by
topologystudentga
on
21 May 2004 04:01 PDT
I apologize, it occured to me yesterday evening that Brower was not
being applied to A, but to the composition. You have been a huge
help. I regret that I did not have the intuition to try any n1 disk,
but I love this proof.

Clarification of Question by
topologystudentga
on
21 May 2004 04:02 PDT
Thanks again for all of your help!
