# Eigenvalues and Eigenvectors

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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

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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.

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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.

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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.

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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 by hkus10
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What is the determinant of an invertible matrix not equal to? What is the relation between the determinant and the eigenvalues of a matrix?

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a) This is what I get let n=lambda.

Since r is an eigenvalue of L, Lx=nx.

What?? If, as you claim $L:V \to V, \ \ x \in V$. then the equation $Lx = nx$ defines $n$ as an eigenvalue and $x$ as its associated eigenvector. Where does $r$ enter the picture?

Since the transformation is invertible, (L^-1)Lx=(L^-1)nx.

I have no idea how you got this. Try $L^{-1}(L(x)) = x$ 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 $x = L^{-1}nx?$

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

$Lx = nx$ defines the eigenvalues(s) $n$ for this operator acting on this vector

$L^{-1}x =rx$ defines the eigenvalue(s) $r$ 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 $r$ and $n$ such that your rearrangement (cleanly done by factorization) makes sense.

Good luck!

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