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Thread: 2-Norm and Cholesky decomposition

  1. #1
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    2-Norm and Cholesky decomposition

    Hello everybody,

    i try to prove this statement:

    if a Matrix M has a cholesky decomposition M=L*L^t, then the equation holds:
    (\| M\|_2) = (\| L\|_2)^2.

    Do you know, how i can prove this?
    I Try this: (\| M\|_2)^2 = \lambda_{max} (M^t * M)=\lambda _{max} (M^2)=(\lambda _{max} (M))^2<br />
    therefore \|M \|_2 = \lambda_max (M)=\lambda_{max} (L*L^t)=\|L^t\|.

    but why is \|L^t\| = \|L\|??

    is this correct? and why the last equation holds?

    Regards.
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  2. #2
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    we have also the decomposition M= B*D*B, whereas D is a diagonal matrix. But i don't see the solution
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  3. #3
    Super Member girdav's Avatar
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    The eigenvalues of LL^t and L^tL are same (in general the eigenvalues of AB and BA are the same).
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    Ok, thanks for your help!! How can i prove your hint'?
    is this correct?:

    AB=A*B*A*A^(-1), that is AB and BA are äquivalent matrices.

    but therefore they have the same eigenvalues:

    since AB-\lambda* I = A*(BA-\lambda*I)*A^(-1) => det(AB-\lambda*I)=det(BA-\lambda*I)

    is this prove correct?

    Thanks a lot.
    Last edited by Sogan; Jun 5th 2011 at 05:05 AM.
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  5. #5
    Super Member girdav's Avatar
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    The result is clear if for example A is invertible. If neither A nor B are invertible, then put A_n :=A+\frac 1n I, where I is the identity matrix. For n sufficiently large, A_n is invertible. Now, use the fact that the determinant is continuous.
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  6. #6
    Super Member girdav's Avatar
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    Quote Originally Posted by Sogan View Post
    Ok, thanks for your help!! How can i prove your hint'?
    is this correct?:

    AB=A*B*A*A^(-1), that is AB and BA are äquivalent matrices.

    but therefore they have the same eigenvalues:

    since AB-\lambda* I = A*(BA-\lambda*I)*A^(-1) => det(AB-\lambda*I)=det(BA-\lambda*I)

    is this prove correct?

    Thanks a lot.
    It's correct if A is invertible.
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    Ok Thank you very much.
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  8. #8
    Super Member girdav's Avatar
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    Take \lambda \in\mathbb C. We have, since a_n is invertible for n\geq n_0:
    \det(AB-\lambda I ) =\lim_{n\to +\infty}\det(A_nB-\lambdaI)=\lim_{n\to+\infty}\det(BA_n-\lambda I)=\det(BA-\lambda I).
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