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Math Help - Eigenvalue Problem

  1. #1
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    Eigenvalue Problem

    A is a simetric metrices nxn. so v\in R^n and v\neq 0
    so (\lambda I -A)^2 v=0 for some \lambda

    prove that for the same v
    (\lambda I -A)v =0

    how i tried to solve it:
    i just collected data from the given.
    simetric matrices is diagonizable.
    B=(\lambda I -A)
    we were given that B^2v=0
    so B^2v \bullet v=0 (dot product is also v)
    so v is orthogonal to B^2v

    what to do now?
    Last edited by transgalactic; June 27th 2011 at 10:24 AM.
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  2. #2
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    Re: Eigenvalue Problem

    The condition specified in the problem does not say anything about v.
    Can you post the question exactly as it was given to you?
    Last edited by Isomorphism; June 27th 2011 at 10:31 AM.
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  3. #3
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    Re: Eigenvalue Problem

    Quote Originally Posted by Isomorphism View Post
    The condition specified in the problem does not say anything about v.
    Can you post the question [COLOR="rgb(0, 0, 0)"]exactly[/COLOR] as it was given to you?
    v is some vector in R^n which solves both homogeneus systems
    i forgot to put v
    in the square system
    Last edited by transgalactic; June 27th 2011 at 10:25 AM.
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  4. #4
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    Re: Eigenvalue Problem

    The solution goes along the lines of:

    (\lambda I - A)^2 v = 0 \implies v^T (\lambda I - A)^2 v = 0 \implies ||(\lambda I - A)v||^2 = 0 \implies (\lambda I - A)v = 0
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  5. #5
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    Re: Eigenvalue Problem

    cant understand why you add tranposed vector
    and going to norm
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  6. #6
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    Re: Eigenvalue Problem

    for any 0-vector (nx1 matrix) u (= 0), and any other row-vector w^T (1xn matrix),

    w^Tu = w^T0 = 0.

    we are given that the vector (\lambda I - A)^2v = 0 for some \lambda.

    so multiplying it by v^T will also give a 0 1x1 matrix, that is, the number 0.

    remember, A is symmetric so λI - A is symmetric too: (\lambda I - A)^T = \lambda I - A.

    thus 0 = v^T(\lambda I - A)^2v = v^T(\lambda I - A)^T(\lambda I - A)v

    = ((\lambda I - A)v)^T(\lambda I - A)v) =\ <(\lambda I - A)v,(\lambda I - A)v>\ = ||(\lambda I - A)v||^2.

    you don't need to pass to the norm, you can just use the positive-definiteness of the inner product.
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  7. #7
    Lord of certain Rings
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    Re: Eigenvalue Problem

    There is no answer to 'why' generally. If you did not understand 'how' any of the implications are true, then I can help you.

    You can do the same proof in another way. Start with the norm step and then expand it. Is that meaningful?
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  8. #8
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    Re: Eigenvalue Problem

    Quote Originally Posted by transgalactic View Post
    A is a simetric metrices nxn. so v\in R^n and v\neq 0
    so (\lambda I -A)^2 v=0 for some \lambda

    prove that for the same v
    (\lambda I -A)v =0

    how i tried to solve it:
    i just collected data from the given.
    simetric matrices is diagonizable.
    B=(\lambda I -A)
    we were given that B^2v=0
    so B^2v \bullet v=0 (dot product is also v)
    so v is orthogonal to B^2v

    what to do now?
    this method is very hard for me.

    i asked my prof for a tip,and he said that i should find the diagonolized form of \lambda I-A
    p is simetric so its diagonisable

    D- \lambda I= P^-1( A-\lambda I) P

    and then i should put it in the equation i need to prove
    (\lambda I -A)v =0
    Last edited by transgalactic; June 28th 2011 at 07:05 AM. Reason: gfnfgn
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  9. #9
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    Re: Eigenvalue Problem

    Quote Originally Posted by Deveno View Post
    for any 0-vector (nx1 matrix) u (= 0), and any other row-vector w^T (1xn matrix),

    w^Tu = w^T0 = 0.

    we are given that the vector (\lambda I - A)^2v = 0 for some \lambda.

    so multiplying it by v^T will also give a 0 1x1 matrix, that is, the number 0.

    remember, A is symmetric so λI - A is symmetric too: (\lambda I - A)^T = \lambda I - A.

    thus 0 = v^T(\lambda I - A)^2v = v^T(\lambda I - A)^T(\lambda I - A)v

    = ((\lambda I - A)v)^T(\lambda I - A)v) =\ <(\lambda I - A)v,(\lambda I - A)v>\ = ||(\lambda I - A)v||^2.

    you don't need to pass to the norm, you can just use the positive-definiteness of the inner product.
    ok but there is a transformation from zero vector into 1x1 zero vector
    its not the same.
    ??


    why
    w^Tu = w^T0 = 0
    ????
    we are not given that u is zero

    then we got to the step of
    ||(\lambda I - A)v||^2
    how to prove that for the same v
    (\lambda I - A)v=0
    from the norm
    ?
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  10. #10
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    Re: Eigenvalue Problem

    in your original post, you said it was given that:

    (\lambda I - A)^2v = 0.

    the whole expression " (\lambda I - A)^2v" can be our u, and we are going to multiply it by the matrix v^T.

    then we re-arrange the terms for our matrix product, to see that it is just an inner product of (\lambda I - A)v with itself.

    the inner product of any vector with itself is positive...unless, our vector (\lambda I - A)v is the 0-vector,

    which is what we hope to prove.
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  11. #11
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    Re: Eigenvalue Problem

    There is another way to proof.

    \vec{v}\neq 0,\;(\lambda I_{n} - A)^2\vec{v}=0\rightarrow \det[(\lambda I_{n} - A)^2]=\det(\lambda I_{n} - A)^2=0\rightarrow \det(\lambda I_{n} - A)=0 \rightarrow \exists\vec{w},\; (\lambda I_{n} - A)\vec{w}=0

    So this method fits for (\lambda I_{n} - A)^{n}\vec{v}=0
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