Eigenvalues for an Invertible Matrix

[Doesy]

Homework Statement

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

Homework Equations

Ax=$$\lambda$$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.

 

[Cyosis]

Let $\mu$ be the eigenvalue of $A^{-1}$. Now multiply the eigenvalue equation with $A^{-1}$ and find $\mu$.

 

[Doesy]

Do you mean I should have

A$$^{2}$$x = A$$\lambda$$x

and

A$$^{-2}$$x = A$$^{-1}$$$$\mu$$x

?

How can I use this to show my Answer? Or do I substitute this second equation into the first?

 

[Cyosis]

No with multiplying the eigenvalue equation I meant $A^{-1}(Ax)=A^{-1}\lambda x$. On a side note, use [tex] brackets around your entire equation not just lambda or mu.

 

[Doesy]

So I can re-arrange as
$$\frac{A^{-1}(A\chi)}{\lambda} = \mu\chi$$
Is that right?

 

[Cyosis]

Sure, because the problem says $\lambda \neq 0$.What is $A^{-1}A$?

 

[Doesy]

Is A^-1*A the Inverse Matrix I?

 

[Cyosis]

Yes it's the identity matrix, which has itself as an inverse. What is Ix?

 

[Doesy]

Inverse multiplied by the eigenvector will give us the original matrix?

 

[Cyosis]

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. $A^{-1}(Ax)=A^{-1}\lambda x \Rightarrow Ix=\lambda \mu x$. Now solve for $\mu$.

 

[Doesy]

Oops, my bad, I meant Identity.

Ix gives us x again.

So:

$$\frac{I\chi}{\lambda\chi} = \mu$$

Is that right? and Ix = x?

 

[Cyosis]

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, $Ix=\lambda \mu x \Rightarrow x=\lambda \mu x \Rightarrow 1*x=\lambda \mu x$. So what is $\mu$?

 

[Doesy]

$$\mu$$ = $$\frac{1}{\lambda}$$

Correct?

 

[Cyosis]

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.

 

[Doesy]

Alright Cool, Here Goes!

$$A\chi = \lambda\chi$$

$$A^{-1}\chi = \lambda^{-1}\chi$$

Let $$\lambda^{-1} = \mu$$

$A^{-1}(Ax)=A^{-1}\lambda x$

We can re-write this as

$$\frac{A^{-1}(A\chi)}{\lambda} = \mu\chi$$

$A^{-1}A = I$

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:

$$\frac{I\chi}{\lambda\chi} = \mu$$

This equates to:

$$\frac{1}{\lambda\chi}$$

This shows that the vector $$\chi$$ is an eigen vector for the Inverse Matrix of A as it's Eigen Value is also the Inverse of $$\chi$$

 

[Cyosis]

Alright Cool, Here Goes!

$$A\chi = \lambda\chi$$

$$A^{-1}\chi = \lambda^{-1}\chi$$

Let $$\lambda^{-1} = \mu$$

You need to proof that $\mu=\lambda^{-1}$, not start with it.

What we are given is the eigenvalue equation $Ax=\lambda x$ with $\lambda$ being an eigenvalue of A. We want to proof that x is an eigenvector of $A^{-1}$ with eigenvalue $\lambda^{-1}$.
So lets multiply the eigenvalue equation by $A^{-1}$.
This yields
$$A^{-1}(Ax)=A^{-1} \lambda x \Rightarrow (A^{-1}A)x=\lambda A^{-1} x \Rightarrow Ix=\lambda A^{-1}x \Rightarrow 1*x=\lambda A^{-1} x \Rightarrow \lambda^{-1}x=A^{-1}x$$.
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 $A^{-1}$ and $\lambda^{-1}$ must be an eigenvalue of $A^{-1}$. 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.

 

[Doesy]

Ohhhh!

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