Special case of eigenvectors/values

[Axioms]

Attached is a question about a special case of eigenvector/values.

 

I am not sure how to solve this and i cant find it in any of my text books or on the web.

 

Any assistance will be greatly appreciated.

 

This is what i have done.

 

let L denote lambda

 

Av=Lv

 

thus Av - Lv = 0

 

(A-L)v=0

 

I try to square both sides here and I do eventually end up with the desired answer. But I have no idea if what I am doing is correct or not. Im just stabbing in the dark here.

 

[studiot]

Well you seem to be on the right track, but it is difficult to say for sure without knowing the question?

 

An important thing with matrices is to make sure you are consistent with premultiplication or postmultiplication since for matrices

 

AB is usually not the same as BA

[Axioms]

Sorry for some reason the question never uploaded with the post: here it is:

 

Let A be an nxn matrix with eigenvalue lambda(L) and associated eigenvector v. Use the definition (Av = Lv)to show that L^2 is an eigenvalue of A^2.

Note: This is a special case of L^k is an eigenvalue A^k.

Thanks for the reply =)

[studiot]

If Av = Lv

 

What does A(Av) = A(Lv) = ?

Edited August 16, 2013 by studiot

[Axioms]

I'm not sure where you are going with this. Sorry, can you explain a bit more?

[studiot]

You want A² (v) yes?

 

Well that is A( A(v) ) isn't it?

 

Substitute for A(v) from your definition as instructed to get my expression in post #4.

 

Since this is homework I am trying to prompt you to conclude the line using the fact the lambda is a scalar.

[Axioms]

Av=Lv

 

A(Av)=A(Lv) multiply both sides by A

 

A(Av)=A(Av) substitute Lv = Av??

 

haha I'm still not getting that eurika moment.

 

-----------------------------------------------------------------------------

 

This is what I tried but I am pretty sure it is wrong:

 

Av=Lv

(A-IL)v=0

(A-IL)(A+IL)=A^2 -IL+IL-L^2=0

A^2=L^2

 

-----------------------------------------------

 

I'm just not sure how to get the L^2 = A^2 answer with the way I understand your explination. I see the A^2 but dont know where you find the L^2. I guess I am just struggling with the principal idea of what they want.

[studiot]

Av = Lv, where L is a scalar.

 

A(Av) = A(Lv) by above definition.

 

Now for any matrix M and any vector w and any scalar S we can interchange the order of the matrix and the scalar

 

so M(Sw) = S(Mw)

 

Can you do it now?

Edited August 16, 2013 by studiot

[Axioms]

oh ok I see. Thanks =].

[studiot]

Glad the penny has dropped.

 

 

Hopefully you can see what the original question meant when it said you can extend this as many times as you like to higher powers.