This problem deals uith a numerical recipe for finding the dominant eigen- value

Question

This problem deals uith a numerical recipe for finding the dominant eigen- value Consider the linear operator on C([O, 1]) given by 1. What is the (minimum) domain of definition of K? 2. Convert the operator on continuous functions to an operator on some finite-dimensional space R" by choosing n equidistant points the inter- val [0, 1] and approximating the choice. The result should be a linear operator Lu = kru for u E Rn and k E R"xn. Give the entries of k in terms of K. Give an expression of r in terms of R. In particular, what are the dimensions of r? integral by a (simple) method of your 3. Find and quote the Perron-Frobenius theorem and list the conditions on K and R so that the theorem applies to your operator L

Explanation / Answer

3T has a positive (real) eigenvalue max such that all other eigenvalues of T satisfy || max. 2 Furthermore max has algebraic and geometric multiplicity one, and has an eigenvector x with x > 0. 3 Any non-negative eigenvector is a multiple of x. 4 More generally, if y 0, y 6= 0 is a vector and µ is a number such that Ty µy then y > 0, and µ max with µ = max if and only if y is a multiple of x. If 0 S T, S 6= T then every eigenvalue of S satisfies || < max. In particular, all the diagonal minors T(i) obtained from T by deleting the i-th row and column have eigenvalues all of which have absolute value < max. If T is primitive, then all other eigenvalues of T satisfy || < max. We now embark on the proof of this important theorem. . Eigenvalue sensitivity and reproductive value. Let P := (I + T) k where k is chosen so large that P is a positive matrix. Then v w, v 6= w Pv < Pw. Recall that Q denotes the positive orthant and that C denotes the intersection of the unit sphere with the positive orthant. For any z Q let L(z) := max{s : sz Tz} = min 1in,zi 6=0 (Tz)i zi . By definition L(rz) = L(z) for any r > 0, so L(z) depends only on the ray through z. If z y, z 6= y we have Pz < Py. Also PT = TP. So if sz Tz then sPz PTz = TPz so L(Pz) L(z).. Eigenvalue sensitivity and reproductive value. L(z) := max{s : sz Tz} = min 1in,zi 6=0 (Tz)i zi . L(Pz) L(z). Furthermore, if L(z)z 6= Tz then L(z)Pz < TPz. So L(Pz) > L(z) unless z is an eigenvector of T with eigenvalue L(z). This suggests that we look for a positive vector which maximizes L.

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