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Math Help - Vector valued DE system

  1. #1
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    Vector valued DE system

    In the last equal sign of the last line below, why is - \lambda z = \ddot{z}? Thanks.

    Given the DE system
    \ddot{\mathbf{x}}=-kA\mathbf{x}

    where \mathbf{x} is n-vector,
    A is an nxn matrix, and
    \mathbf{v} is a an eigenvector of A,

    let \mathbf{x}=z\mathbf{v}, then

    \ddot{\mathbf{x}}=-kA\mathbf{x}=-kA(z\mathbf{v})=-k(A\mathbf{v})z=-k(\lambda\mathbf{v})z=k(-\lambda z)\mathbf{v}=k\ddot{z}\mathbf{v}.
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  2. #2
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    Re: Vector valued DE system

    Quote Originally Posted by MSUMathStdnt View Post
    In the last equal sign of the last line below, why is - \lambda z = \ddot{z}? Thanks.

    Given the DE system
    \ddot{\mathbf{x}}=-kA\mathbf{x}

    where \mathbf{x} is n-vector,
    A is an nxn matrix, and
    \mathbf{v} is a an eigenvector of A,

    let \mathbf{x}=z\mathbf{v}, then

    \ddot{\mathbf{x}}=-kA\mathbf{x}=-kA(z\mathbf{v})=-k(A\mathbf{v})z=-k(\lambda\mathbf{v})z=k(-\lambda z)\mathbf{v}=k\ddot{z}\mathbf{v}.
    Look back to the first of the line. What the first to the next to last term shows is that k(-\lamba z)\mathbf{v}= \ddot{\mathbf{x}} = \ddot{\mathbf{zv}}=  k\ddot{z}v because v is a constant.
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  3. #3
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    Re: Vector valued DE system

    Quote Originally Posted by HallsofIvy View Post
    Look back to the first of the line. What the first to the next to last term shows is that k(-\lamba z)\mathbf{v}= \ddot{\mathbf{x}} = \ddot{\mathbf{zv}}=  k\ddot{z}v because v is a constant.
    I'm not sure what you're trying to tell me. I understand that \mathbf{v} is constant. But how does what you wrote mean that - \lambda z = \ddot{z}?

    Is it because - \lambda z = -\lambda I_n z, ( I being the identity matrix), which would make the right side of this into the left side of the given system with \lambda as k and I as A?
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  4. #4
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    Re: Vector valued DE system

    Yes,. if for any (non-zero) vector, v, av= bv, then a= b.
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