Are Both Matrices Diagonalizable?

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Homework Help Overview

The discussion revolves around the diagonalizability of two matrices, focusing on their eigenvalues and eigenvectors. Participants are examining the conditions under which a matrix can be diagonalized, particularly in the context of real numbers.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss solving for eigenvectors and eigenvalues, questioning the correctness of these calculations. There are inquiries about the characteristic equation and the relationship between algebraic and geometric multiplicities. Some participants suggest showing work for collaborative assistance.

Discussion Status

The discussion is active, with various interpretations of the diagonalizability of the matrices being explored. Some participants assert that the matrices are diagonalizable, while others express doubts based on their findings of eigenvalues and eigenvectors. Guidance has been offered regarding the need for distinct eigenvalues and the implications of having insufficient linearly independent eigenvectors.

Contextual Notes

There is an emphasis on considering diagonalizability over the reals, which affects the conclusions drawn about the matrices. Participants note the importance of having three distinct real eigenvalues and corresponding eigenvectors for diagonalizability.

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Homework Statement


Are the following matrices diagonalizable:

[PLAIN]http://img693.imageshack.us/img693/4198/91350081.jpg

The Attempt at a Solution



I solved for the eigenvectors of both matrices, but only found one eigenvector for each. This means that neither of the two matrices have at least two linearly independent eigenvectors, which means neither are diagonalizable.

Am I correct?
 
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show your working & people can check it rather than having to do the whole problem (i haven't checked it)

however you are correct that for a 3x3 if the sum of the dimensions of the eigenspaces is <3, the matrix is not diagonalisable
 
You must not be solving for the eigenvectors correctly, because they're both diagonalizable.
 
vela said:
You must not be solving for the eigenvectors correctly, because they're both diagonalizable.

For a) did you get an eigenvalue of 3 and for b) 0, 1, and 3?
 
sonce again - show your working, and i'll try & help, but I won't do the whole problem for you ;)

a) I'm not so convinced this one is diagonalisable - as a start, what is the characteristic equation & what is the multiplicity for the eigenvalue \lambda = 3 (consider both algerabraic & geometric multiplicity) -

b) if you have 3 distinct real eigenvectors, you should be able to find 3 corresponding distinct real eigenvectors - what does this tell you ablout the diagonalisability of your matrix?
 
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I got three distinct eigenvalues for the first one as well. Two are imaginary, but that's ok.
 
I assume the question means diagonalisable over the reals though, so that result would imply the matrix in a) is not diagonalisable...
 
lanedance said:
I assume the question means diagonalisable over the reals though, so that result would imply the matrix in a) is not diagonalisable...

That would be true, if that's what temaire means.
 
Dick said:
That would be true, if that's what temaire means.

Yes, we're only considering real numbers here. So for a) I found only one real eigenvalue, 3, which doesn't have at least 3 linearly independent eigenvectors, so it's not diagonalizable. For b) there are three real eigenvalues, 0, 1, and 3, and I was only able to come up with one linearly independent eigenvector, (-1, 1, 1) for the eignevector 0. For eigenvectors 1 and 3, I get no solutions, so therefore no eigenvectors. Did I go wrong here?
 
  • #10
Yes, you can find eigenvectors for each of the eigenvalues.
 

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