Solving Eigenvalue Problems: What Is Induction?

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Homework Help Overview

The discussion revolves around solving eigenvalue problems, specifically focusing on the properties of eigenvalues and eigenvectors when applied to powers of a matrix. The original poster seeks guidance on how to approach the problem and whether their reasoning is correct.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between eigenvalues and powers of matrices, questioning how to formally prove the general case using induction. There is a discussion on whether the reasoning presented is sufficient or if a more formal proof is necessary.

Discussion Status

Some participants have provided guidance on using induction to formalize the reasoning. There is an ongoing exploration of the concept of induction itself, with some expressing uncertainty about its application and others suggesting it as a necessary approach for completeness.

Contextual Notes

One participant mentions a lack of familiarity with induction, indicating that it may not have been covered in their coursework. This raises questions about the expectations for the problem and the level of detail required in the solution.

temaire
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Homework Statement


[PLAIN]http://img28.imageshack.us/img28/5227/79425145.jpg


The Attempt at a Solution



I'm not exactly sure how to go about this problem. How do I start?
 
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Start with A^2. A^2(x)=A(A(x)). What's that in terms of lambda?
 


Is it lambda^2(x)
 


temaire said:
Is it lambda^2(x)

Sure. So that means x is an eigenvector of A^2 with eigenvalue lambda^2, right? The statement for general N>0 follows in the same way.
 


So the solution is simply:
Ax = lambda x
Therefore A^n x = lambda^n x ?

Is there something I'm missing?
 


temaire said:
So the solution is simply:
Ax = lambda x
Therefore A^n x = lambda^n x ?

Is there something I'm missing?

That's what the problem is asking you to show, isn't it?
 


Dick said:
That's what the problem is asking you to show, isn't it?

So what you showed me with A^2 is all I need to answer the question?
 


temaire said:
Yes it is, but is that all there is to it?

If you understand why it's true, then yes, that's all there is to it. If you want to be formal about proving it you might want to present it as an induction proof.
 


Here is my solution:

A^2(x) = A(A(x))
A^2(x) = lambda^2(x)

Therefore x is an eigenvector of A^2 with eigenvalue lambda^2. The general statement for n>0 follows in the same way.

Is this complete?
 
  • #10


You should probably use induction rather than say "The general statement for n>0 follows in the same way." That's very vague.
 
  • #11


Mark44 said:
You should probably use induction rather than say "The general statement for n>0 follows in the same way." That's very vague.

What do you mean by induction? I've never learned it.
 
  • #13


temaire said:
What do you mean by induction? I've never learned it.

Induction is the formal way to show it. If you've never heard of it and aren't expected to use it the alternative is 'hand waving'. It's easy enough to prove A^3(x)=lambda^3 x using A^2(x)=lambda^2 x. That makes it easy to show A^4(x)=lambda^4 x, etc, etc. So we see it's true for all n. Induction is just the formal way to state 'etc etc'. It's up to you to decide whether the course requires it.
 

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