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Eigenvalues for an Invertible Matrix

  • Thread starter Doesy
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  • #1
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Homework Statement


A is an invertible matrix, x is an eigenvector for A with an eiganvalue [tex]\lambda[/tex] [tex]\neq[/tex]0 Show that x is an eigenvector for A^-1 with eigenvalue [tex]\lambda[/tex]^-1


Homework Equations


Ax=[tex]\lambda[/tex]x
(A - I)x

The Attempt at a Solution



I know that I need to find x and then apply to the inverses of my Matrix and eigenvalue, but how do I know what matrix to use for A? Do I use the inverse matrix as it is an invertible matrix? Can I use any invertible matrix to prove this?

Thanks in advance.
 

Answers and Replies

  • #2
Cyosis
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Let [itex]\mu[/itex] be the eigenvalue of [itex]A^{-1}[/itex]. Now multiply the eigenvalue equation with [itex]A^{-1}[/itex] and find [itex]\mu[/itex].
 
  • #3
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Do you mean I should have

A[tex]^{2}[/tex]x = A[tex]\lambda[/tex]x

and

A[tex]^{-2}[/tex]x = A[tex]^{-1}[/tex][tex]\mu[/tex]x

?

How can I use this to show my Answer? Or do I substitute this second equation into the first?
 
  • #4
Cyosis
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No with multiplying the eigenvalue equation I meant [itex]A^{-1}(Ax)=A^{-1}\lambda x[/itex]. On a side note, use [tex] brackets around your entire equation not just lambda or mu.
 
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  • #5
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So I can re-arrange as
[tex]\frac{A^{-1}(A\chi)}{\lambda} = \mu\chi[/tex]
Is that right?
 
  • #6
Cyosis
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Sure, because the problem says [itex]\lambda \neq 0 [/itex].What is [itex]A^{-1}A[/itex]?
 
  • #7
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Is A^-1*A the Inverse Matrix I?
 
  • #8
Cyosis
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Yes it's the identity matrix, which has itself as an inverse. What is Ix?
 
  • #9
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Inverse multiplied by the eigenvector will give us the original matrix?
 
  • #10
Cyosis
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Are you calling I the inverse? It's the Identity matrix. If you let the Identity matrix work on a vector it yields that vector without making any changes to it. We're moving in the wrong direction now so lets start at the eigenvalue equation again. [itex] A^{-1}(Ax)=A^{-1}\lambda x \Rightarrow Ix=\lambda \mu x[/itex]. Now solve for [itex]\mu[/itex].
 
  • #11
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Oops, my bad, I meant Identity.

Ix gives us x again.

So:

[tex]\frac{I\chi}{\lambda\chi} = \mu[/tex]

Is that right? and Ix = x?
 
  • #12
Cyosis
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Well x are vectors how do you define a vector divided by another vector? While doing it in this case and letting the xs cancel you will get the correct answer, I would suggest you don't do this on your exam.

This is what you should do, [itex]Ix=\lambda \mu x \Rightarrow x=\lambda \mu x \Rightarrow 1*x=\lambda \mu x[/itex]. So what is [itex]\mu[/itex]?
 
  • #13
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[tex]\mu[/tex] = [tex]\frac{1}{\lambda}[/tex]

Correct?
 
  • #14
Cyosis
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Yes, perhaps it's nice if you write the full proof down now so we can see if you don't do any operations that you shouldn't really be doing.
 
  • #15
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Alright Cool, Here Goes!

[tex]A\chi = \lambda\chi[/tex]

[tex]A^{-1}\chi = \lambda^{-1}\chi[/tex]

Let [tex]\lambda^{-1} = \mu[/tex]

[itex]
A^{-1}(Ax)=A^{-1}\lambda x
[/itex]

We can re-write this as

[tex]
\frac{A^{-1}(A\chi)}{\lambda} = \mu\chi
[/tex]

[itex]
A^{-1}A = I
[/itex]

Here I is the Identity Matrix

Ix = x

As The Identity matrix multiplied by a vector value does not change the position of the vector. So we now have:

[tex]
\frac{I\chi}{\lambda\chi} = \mu
[/tex]

This equates to:

[tex]\frac{1}{\lambda\chi}[/tex]

This shows that the vector [tex]\chi[/tex] is an eigen vector for the Inverse Matrix of A as it's Eigen Value is also the Inverse of [tex]\chi[/tex]
 
  • #16
Cyosis
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Alright Cool, Here Goes!

[tex]A\chi = \lambda\chi[/tex]

[tex]A^{-1}\chi = \lambda^{-1}\chi[/tex]

Let [tex]\lambda^{-1} = \mu[/tex]
You need to proof that [itex]\mu=\lambda^{-1}[/itex], not start with it.

What we are given is the eigenvalue equation [itex]Ax=\lambda x[/itex] with [itex]\lambda[/itex] being an eigenvalue of A. We want to proof that x is an eigenvector of [itex]A^{-1}[/itex] with eigenvalue [itex]\lambda^{-1}[/itex].
So lets multiply the eigenvalue equation by [itex]A^{-1}[/itex].
This yields
[tex]A^{-1}(Ax)=A^{-1} \lambda x \Rightarrow (A^{-1}A)x=\lambda A^{-1} x \Rightarrow Ix=\lambda A^{-1}x \Rightarrow 1*x=\lambda A^{-1} x \Rightarrow \lambda^{-1}x=A^{-1}x[/tex].
A matrix working on a vector that yields the same vector multiplied by a scalar is an eigenvalue equation. So x must be an eigenvector of [itex]A^{-1}[/itex] and [itex]\lambda^{-1}[/itex] must be an eigenvalue of [itex]A^{-1}[/itex]. There is no dividing of any vectors happening here! If you want to divide two vectors you must first define how division with vectors works.
 
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  • #17
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Ohhhh!

Thanks heaps man, is there anyway I can +rep you or something?
 

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