Eigenvalues for an Invertible Matrix

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].

 

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.

 

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

 

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

 

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].

 

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]?

 

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.

 

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.