Find Eigenvalue/Eigenvector

  • #1
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779

Homework Statement



If invertible matrices A and B have eigenvalues α and β resp. wrt to a common eigenvector x, what is the eigenvalue of

4A3B-4-17AB2 + ∏I

(We'll call this equation (1))

Homework Equations



Ax = αx
Bx = βx


The Attempt at a Solution



I think I'm ok for the actual eigenvalue. Basically we "exchange" α for A and β for B in the equation above, since Ax = αx implies A2x = α2x and so on and so forth. ∏I is just an identity matrix multiplied by ∏, so it's eigenvalue is ∏ (with multiplicity two).

So I get 4α24 - 17αβ2 + ∏

But I can't seem to figure out how to approach getting the corresponding Eigenvector. I know the characteristic equation

det(A - λI) = 0

may tell me something, but if I plug in (1) for A and the eigenvalue above for λ I just get a complete mess.

-Dave K
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619

Homework Statement



If invertible matrices A and B have eigenvalues α and β resp. wrt to a common eigenvector x, what is the eigenvalue of

4A3B-4-17AB2 + ∏I

(We'll call this equation (1))

Homework Equations



Ax = αx
Bx = βx


The Attempt at a Solution



I think I'm ok for the actual eigenvalue. Basically we "exchange" α for A and β for B in the equation above, since Ax = αx implies A2x = α2x and so on and so forth. ∏I is just an identity matrix multiplied by ∏, so it's eigenvalue is ∏ (with multiplicity two).

So I get 4α24 - 17αβ2 + ∏

But I can't seem to figure out how to approach getting the corresponding Eigenvector. I know the characteristic equation

det(A - λI) = 0

may tell me something, but if I plug in (1) for A and the eigenvalue above for λ I just get a complete mess.

-Dave K

You can't really find the eigenvector unless you have actual values for the A and B matrices. Do you?
 
  • #3
I like Serena
Homework Helper
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176
Hi Dave!

Let's call C = 4A3B-4-17AB2 + ∏I
Then, as you said, Cx = (4α24 - 17αβ2 + ∏)x

Let's call λ=4α24 - 17αβ2 + ∏.
Then Cxx.

You had already found the eigenvalue λ.
What do you think the corresponding eigenvector is? :wink:
 
  • #4
1,047
779
Hi Dave!

Let's call C = 4A3B-4-17AB2 + ∏I
Then, as you said, Cx = (4α24 - 17αβ2 + ∏)x

Let's call λ=4α24 - 17αβ2 + ∏.
Then Cxx.

You had already found the eigenvalue λ.
What do you think the corresponding eigenvector is? :wink:

I did try that but I'm not sure how to "solve" for x. As the other poster pointed out, I don't even know what the values of C are... I don't even know what dimension so I can't use some identity matrix and use λ (the mess) as a scalar... Not sure what I'm supposed to do.
 
  • #5
Dick
Science Advisor
Homework Helper
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I did try that but I'm not sure how to "solve" for x. As the other poster pointed out, I don't even know what the values of C are... I don't even know what dimension so I can't use some identity matrix and use λ (the mess) as a scalar... Not sure what I'm supposed to do.

You can't solve for x any more than you can solve for α and β. x is the eigenvector. I think you are done.
 
  • #6
I like Serena
Homework Helper
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176
I did try that but I'm not sure how to "solve" for x. As the other poster pointed out, I don't even know what the values of C are... I don't even know what dimension so I can't use some identity matrix and use λ (the mess) as a scalar... Not sure what I'm supposed to do.

The vector x is given as the common eigenvector of A and B.
None of them is specified other than that they are given to exist.
C has the same common eigenvector x as A and B.
 
  • #7
1,047
779
You can't solve for x any more than you can solve for α and β. x is the eigenvector. I think you are done.

Yeah, we went over this in class, and the eigenvector is just x. Ok then. Thanks everyone.
 

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