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## Homework Statement

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

## The Attempt at a Solution

I'm not exactly sure how to go about this problem. How do I start?

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- Thread starter temaire
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[PLAIN]http://img28.imageshack.us/img28/5227/79425145.jpg [Broken]

I'm not exactly sure how to go about this problem. How do I start?

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- #2

Dick

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Start with A^2. A^2(x)=A(A(x)). What's that in terms of lambda?

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Is it lambda^2(x)

- #4

Dick

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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.

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So the solution is simply:

Ax = lambda x

Therefore A^n x = lambda^n x ?

Is there something I'm missing?

- #6

Dick

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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?

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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?

- #8

Dick

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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.

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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

Mark44

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

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

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Mark44

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- #13

Dick

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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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