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Math Help - matrix diagonalisation - real and complex matrices - some observations

  1. #16
    A Plied Mathematician
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    I think the problem with your computation is that, while A is real, its eigenvalues are not necessarily real (we have not assumed that A is symmetric). Hence, in order to test for symmetry, you should be computing complex conjugate transposes (I use daggers for that operation); that is, what you must really show is that

    A^{\dagger}=(P^{\dagger}DP)^{\dagger}=P^{\dagger}D  ^{\dagger}P=P^{\dagger}DP=A, but it is not necessarily true that

    D^{\dagger}=D.

    You can prove that every normal matrix is diagonalizable, and it is also true that not every normal matrix is symmetric.
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  2. #17
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Ackbeet View Post
    ... while A is real, its eigenvalues are not necessarily real (we have not assumed that A is symmetric).
    That is right. But look at my previous post, I included in the hyphotesis D\in\mathbb{R}^{n\times n}" . That is, "ortoghonally diagonalizable on \mathbb{R}".


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  3. #18
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    Ok, but I would claim that's an additional hypothesis that's not implied by being orthogonally diagonalizable. Not all orthogonally diagonalizable real matrices have real eigenvalues (which is what your D consists of).
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  4. #19
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Ok, but I would claim that's an additional hypothesis that's not implied by being orthogonally diagonalizable. Not all orthogonally diagonalizable real matrices have real eigenvalues (which is what your D consists of).
    The following solve the question:

    Definition Consider A\in \mathbb{R}^{n\times n} . We say that A is orthogonally diagonalizable if and only if there exists an orthogonal matrix Q\in \mathbb{R}^{n\times n} such that Q^tAP=D\in  \mathbb{R}^{n\times n} with D diagonal.

    I've just used that definition, which implies I've not added any hypothesis. For your comments I see that you are using another definition, I know which one, and I accept it. So, you and me are right and we are not talking about a mathematical problem but about a convention problem .


    Fernando Revilla
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  5. #20
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    Sure, fine. You can define things that way. But then you can't have normal matrices to be orthogonally diagonalizable.
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  6. #21
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Sure, fine. You can define things that way.

    Actually, that is a standard definition.

    But then you can't have normal matrices to be orthogonally diagonalizable.

    We can, A\in\mathbb{R}^{n\times n} symmetric implies A normal.


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  7. #22
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    Quote Originally Posted by FernandoRevilla View Post
    We can, A\in\mathbb{R}^{n\times n} symmetric implies A normal.Fernando Revilla
    I meant the other direction: you'll have some normal matrices that are not symmetric, which you won't be able to orthogonally diagonalize.
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  8. #23
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Ackbeet View Post
    I meant the other direction: you'll have some normal matrices that are not symmetric, which you won't be able to orthogonally diagonalize.
    Right.

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