hkus10 Posted May 5, 2011 Share Posted May 5, 2011 Let L : V>>>V be an invertible linear operator and let lambda be an eigenvalue of L with associated eigenvector x. a) Show that 1/lambda is an eigenvalue of L^-1 with associated eigenvector x. For this question, the things I know are that L is onto and one to one. Therefore, how to prove this question? b) state the analogous statement for matrices. What does "state" the analogous statement mean? Thanks Link to comment Share on other sites More sharing options...

DrRocket Posted May 5, 2011 Share Posted May 5, 2011 Let L : V>>>V be an invertible linear operator and let lambda be an eigenvalue of L with associated eigenvector x. a) Show that 1/lambda is an eigenvalue of L^-1 with associated eigenvector x. For this question, the things I know are that L is onto and one to one. Therefore, how to prove this question? b) state the analogous statement for matrices. What does "state" the analogous statement mean? Thanks a) You should know a bit more that just that L is 1-1 and onto. What is L(x) ? What is L^-1(L(x)) ? b) Writing it down would be considered "stating" it. 1 Link to comment Share on other sites More sharing options...

hkus10 Posted May 6, 2011 Author Share Posted May 6, 2011 b) Writing it down would be considered "stating" it. What does the sentence want me to write down? The key is I do not understand the question. Link to comment Share on other sites More sharing options...

DrRocket Posted May 6, 2011 Share Posted May 6, 2011 What does the sentence want me to write down? The key is I do not understand the question. Use what you know about the correspondence between matrices and linear transformations. 1 Link to comment Share on other sites More sharing options...

hkus10 Posted May 6, 2011 Author Share Posted May 6, 2011 (edited) a) You should know a bit more that just that L is 1-1 and onto. What is L(x) ? What is L^-1(L(x)) ? a) This is what I get let n=lambda. Since r is an eigenvalue of L, Lx=nx. Since the transformation is invertible, (L^-1)Lx=(L^-1)nx. ==> Ix=n(L^-1)x, where I=indentity matrix At this point, I want to divide both sides by r. However, how can I be sure r is not equal to zero? Thanks Edited May 6, 2011 by hkus10 Link to comment Share on other sites More sharing options...

ajb Posted May 6, 2011 Share Posted May 6, 2011 What is the determinant of an invertible matrix not equal to? What is the relation between the determinant and the eigenvalues of a matrix? 1 Link to comment Share on other sites More sharing options...

Xerxes Posted May 7, 2011 Share Posted May 7, 2011 a) This is what I get let n=lambda. Since r is an eigenvalue of L, Lx=nx. What?? If, as you claim [math]L:V \to V, \ \ x \in V[/math]. then the equation [math]Lx = nx[/math] defines [math] n[/math] as an eigenvalue and [math]x[/math] as its associated eigenvector. Where does [math] r[/math] enter the picture? Since the transformation is invertible, (L^-1)Lx=(L^-1)nx. I have no idea how you got this. Try [math]L^{-1}(L(x)) = x[/math] by the simple fact, as given by you that the transformation is bijective ==> Ix=n(L^-1)x, where I=indentity matrix The identity operator/matrix acting on any vector is the vector itself. How can it be that [math]x = L^{-1}nx?[/math] So, I firmly believe that students should do their own homework, but here is a big hint. Assume that you mis-typed, and meant that [math]Lx = nx[/math] defines the eigenvalues(s) [math] n[/math] for this operator acting on this vector [math]L^{-1}x =rx[/math] defines the eigenvalue(s) [math]r[/math] for this operator acting on the same vector. So, first rearrange each of these 2 equalities, and using any, all or none of the above, find a relation between [math]r[/math] and [math]n[/math] such that your rearrangement (cleanly done by factorization) makes sense. Good luck! 1 Link to comment Share on other sites More sharing options...

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