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Q: Eigenvalues ( Answered 5 out of 5 stars,   0 Comments )
Subject: Eigenvalues
Category: Science > Math
Asked by: topologystudent-ga
List Price: $5.00
Posted: 20 May 2004 10:21 PDT
Expires: 19 Jun 2004 10:21 PDT
Question ID: 349444
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 mathtalk-ga on 20 May 2004 11:05 PDT
Rather than apply Brouwer's Fixed Point Theorem to the n-disk here:

you should consider the (n-1)-disk formed by the intersection of the
positive orthant in R^n and the boundary of n-disk.

Show that the mapping induced by centrally projecting Ax onto this
(topological) (n-1)-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 well-defined and continuous.

Let me know if you'd like more detailed discussion.

regards, mathtalk-ga

Clarification of Question by topologystudent-ga 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 (n-1) 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.


Request for Question Clarification by mathtalk-ga 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 (n-1)-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, mathtalk-ga

Request for Question Clarification by mathtalk-ga on 20 May 2004 20:10 PDT
Perhaps I need to make the (n-1)-disk a little clearer.  I was
thinking about the portion of the "unit" n-disk 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 (n-1)-simplex in standard position, ie. the convex
hull of the unit vectors in R^n.  Topologically this is an (n-1)-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, mathtalk-ga

Clarification of Question by topologystudent-ga 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 n-1 disk,
but I love this proof.

Clarification of Question by topologystudent-ga on 21 May 2004 04:02 PDT
Thanks again for all of your help!
Subject: Re: Eigenvalues
Answered By: mathtalk-ga on 21 May 2004 07:24 PDT
Rated:5 out of 5 stars
Hi, topologystudent-ga:

As you noticed, we cannot make much headway in finding an eigenvector
if Brouwer's fixed point theorem:

[Brouwer's Fixed Point Theorem]

is applied to A, eg. acting on the unit ball B^n, because of the
trivial fixed point of any linear mapping.  This suggests the need to
construct a mapping that acts on a subset D of R^n that excludes the

Given that the entries of A are all positive, it seems that focusing
on the positive (or _nonnegative_) orthant will be productive, as A
maps this set into itself.  But to satisfy the hypotheses of Brouwer's
fixed point theorem we need a disk!

A n-dimensional disk might be embedded in this "positive cone", but
the mapping to which we apply the fixed point theorem needs to map D
into itself.  Hence the idea of "flattening" entire cone down to an
(n-1) disk, either the portion D of the unit (n-1) sphere contained in
the positive orthant or (as subsequently outlined) the simplex
generated by the n standard basis vectors (as a convex hull of those n

The critical step in this proof is to show a lower bound on ||Ax||. 
Here it is necessary to make use of the fact that all entries in A are
positive, and not just nonnegative.  [Exercise: Give a nonzero 2x2
nonnegative matrix which has no positive eigenvalue.]  Hint: Consider
the smallest entry in A and the largest entry in x.

Once we know that ||Ax|| on D is always at least c > 0, then we have
no difficulty in arguing that F(x) = Ax/||Ax|| is well-defined and
continuous on D.

regards, mathtalk-ga
topologystudent-ga rated this answer:5 out of 5 stars
Fantastic answer.  I appreciated the discussion before the final
answer.  Very good explaination, alternate method, and exercise. 
Thanks again for all the help.

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