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Non-diagonalisable matrix of a linear trans.

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data

    If the matrix of a linear transformation relative to a basis is non-diagonalisable, then for any other choice of basis, it too will be non-diagonalisable. Prove this is the case.

    2. Relevant equations

    Similar matrices.

    3. The attempt at a solution

    Let APB=[T]BPB, where A is a nxn matrix.

    But I'm stuck. Not sure what to do next.
     
  2. jcsd
  3. Sep 16, 2009 #2

    Office_Shredder

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    What is your definition of diagonalizable and non-diagonalizable? It makes a pretty big difference since there are a couple ways to approach this depending on what we can assume and what we want to prove
     
  4. Sep 16, 2009 #3

    Dick

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    If it's diagonalizable then it has a basis of consisting of n eigenvectors. The choice of the basis to represent the matrix doesn't change that.
     
  5. Sep 16, 2009 #4
    There might be an easier way to show this, but in physics diagonalizing matricies always leads to eigenvalue problems.... I believe your statement implies the matrix doesn't have a closed form for its eigenvalue equation. So
    [tex]|A-\omega I|=0[/tex] wouldn't have n roots and hence can't be diagonalized.

    But as office-shredder says, there is more than one way to diagonalize a matrix...
     
  6. Sep 16, 2009 #5

    Dick

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    If you mean real diagonalizable vs complex diagonalizable, I'm not sure that's what this is about.
     
  7. Sep 17, 2009 #6

    Office_Shredder

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    No, I mean, for example, the definition of non-diagonalizable that I would use, and a lot of other people, is something like "a matrix which is not diagonal under any change of basis". Making the whole question kind of moot. Someone thought this question was worth asking, so obviously there's a different definition floating around.

    That would be a diagonal and a... non-diagonal matrix. What is a diagonalizable matrix, and what is a non-diagonalizable matrix? You can't expect to solve the problem without knowing what you're talking about
     
  8. Sep 17, 2009 #7

    Dick

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    Now you are just confusing me. What does being diagonalizable have to do with the choice of a basis? The problem just says show it's independent of basis. If you state the condition in terms of eigenvectors, it's clearly basis independent.
     
  9. Sep 17, 2009 #8
    Looks like I've caused too much trouble here. Thanks for trying.

    Pibeta.
     
  10. Sep 17, 2009 #9

    Dick

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    Don't leave! Just show that the quality of a matrix being diagonalizable is independent of the basis by stating it in a way that is independent of the basis. I'm not sure what this flurry of confusion was about, but it's nothing about you causing trouble.
     
  11. Sep 17, 2009 #10

    HallsofIvy

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    "Doesn't have a closed form for its eigenvalue equation"? What does that mean?

    That doesn't follow. If an operator (on an n dimensional vector space) has n distinct roots then it is diagonalizable, but if not it still may be diagonalizable. As Dick said, it depends on the number of independent eigenvectors, not eigenvalues.

    No, office-shredder did not say that. He said there is more than one way to define "diagonalizable".
     
  12. Sep 19, 2009 #11
    Thank you for all of this help (I know what diagonalisable means, but didn't realise they was more than one way to define it). And I've decided to come back too.

    Anyways, a hint I was given was to use [T]C=PC<-B[T]B(PC<-B)-1 and prove the contradiction, which is the matrix is diagonalisable.

    To show it is diagonalisable, I got (using the hint) [T]C=PBAPB-1 (where A=PB[T]B(PB)-1 )


    but I'm not sure if this is right and where to go from here...
     
  13. Sep 20, 2009 #12

    HallsofIvy

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    I'm glad to know that you know what "diagonalizable" means- that helps a lot! But to help you we still need to know what your definition of "diagonalizable" is! Please tell us.
     
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