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 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: http://en.wikipedia.org/wiki/Brouwer_fixed_point_theorem 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. Correct?``` 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!`
 ```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] http://en.wikipedia.org/wiki/Brouwer_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 origin. 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 points). 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: ```Fantastic answer. I appreciated the discussion before the final answer. Very good explaination, alternate method, and exercise. Thanks again for all the help.```