Eigenvalues and Eigenvectors

[hkus10]

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

[DrRocket]

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.

[hkus10]

 

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.

[DrRocket]

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.

[hkus10]

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

[ajb]

What is the determinant of an invertible matrix not equal to? What is the relation between the determinant and the eigenvalues of a matrix?

[Xerxes]

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!