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Math Help - Eigenvalues

  1. #16
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    Re: Eigenvalues

    we used row-operations and column operations to evaluate det(A - λI).

    the matrix we wound up with is NOT A-λI, it just has the same determinant.

    it is clear that with your original matrix A(e1) = (1,1,1,.....,1) ≠ n(e1) = (n,0,0,.....,0).
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  2. #17
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    Re: Eigenvalues

    Quote Originally Posted by Deveno View Post
    we used row-operations and column operations to evaluate det(A - λI).

    the matrix we wound up with is NOT A-λI, it just has the same determinant.

    it is clear that with your original matrix A(e1) = (1,1,1,.....,1) ≠ n(e1) = (n,0,0,.....,0).
    What?
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  3. #18
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    Re: Eigenvalues

    you're trying to find the eigenvectors of A = (a_ij) ,where every entry is 1, right?

    if v is an eigenvector, then Av = λv, so (A-λI)v = 0.

    when λ = n, this is: (A-nI)v = 0, a homogeneous linear system of equations.

    e1 is not a solution to that system.
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  4. #19
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    Re: Eigenvalues

    Quote Originally Posted by Deveno View Post
    you're trying to find the eigenvectors of A = (a_ij) ,where every entry is 1, right?

    if v is an eigenvector, then Av = λv, so (A-λI)v = 0.

    when λ = n, this is: (A-nI)v = 0, a homogeneous linear system of equations.

    e1 is not a solution to that system.
    Is this not we obtain when we plug in n?

    \begin{bmatrix}0&1&\cdots & 1\\0&-n&\cdots &0\\ \vdots & 0& \ddots \\ 0&\cdots & &-n \end{bmatrix}\Rightarrow\begin{bmatrix}0&0&\cdots & 0\\0&1&\cdots &0\\ \vdots & 0& \ddots \\ 0&\cdots & &1 \end{bmatrix}
    Last edited by dwsmith; October 15th 2011 at 05:45 PM.
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  5. #20
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    Re: Eigenvalues

    Quote Originally Posted by dwsmith View Post
    Is this not we obtain when we plug in n?

    \begin{vmatrix}0&1&\cdots & 1\\0&-n&\cdots &0\\ \vdots & 0& \ddots \\ 0&\cdots & &-n \end{vmatrix}\Rightarrow\begin{vmatrix}0&0&\cdots & 0\\0&1&\cdots &0\\ \vdots & 0& \ddots \\ 0&\cdots & &1 \end{vmatrix}
    is:
    \begin{bmatrix}0&1&\cdots & 1\\0&-n&\cdots &0\\ \vdots & 0& \ddots \\ 0&\cdots & &-n \end{bmatrix}

    the original matrix we are finding eigenvalues/eigenvectors for? no, it is not. that matrix is:

    \begin{bmatrix}1&1&\cdots & 1\\1&1&\cdots &1\\ \vdots & 1& \ddots \\ 1&\cdots & &1 \end{bmatrix}
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  6. #21
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    Re: Eigenvalues

    Quote Originally Posted by Deveno View Post
    is:
    \begin{bmatrix}0&1&\cdots & 1\\0&-n&\cdots &0\\ \vdots & 0& \ddots \\ 0&\cdots & &-n \end{bmatrix}

    the original matrix we are finding eigenvalues/eigenvectors for? no, it is not. that matrix is:

    \begin{bmatrix}1&1&\cdots & 1\\1&1&\cdots &1\\ \vdots & 1& \ddots \\ 1&\cdots & &1 \end{bmatrix}
    \begin{bmatrix}1-n&1&\cdots & 1\\1&1-n&\cdots &1\\ \vdots & 1& \ddots \\ 1&\cdots & &1-n \end{bmatrix}

    From this matrix, I can obtain the previous one I posted by elementary row operations. This matrix will revert to the matrix in post 5 and from there it can be further simplified to the one I just posted but a 2 in the a_{11} position not a zero.
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  7. #22
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    Re: Eigenvalues

    the matrix in post 5 was not obtained by elementary row operations, but by elementary row AND column operations.

    column operations DO NOT preserve the solution space.
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  8. #23
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    Re: Eigenvalues

    Quote Originally Posted by Deveno View Post
    the matrix in post 5 was not obtained by elementary row operations, but by elementary row AND column operations.

    column operations DO NOT preserve the solution space.
    From simplifying I obtain, (2-n)x_1=-x_3-\cdots -x_n and x_1=x_2 with the rest free variables. What can I do now if this is correct?
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  9. #24
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    Re: Eigenvalues

    is it not the case that:

    \begin{bmatrix}1&1&\cdots&1\\1&1&\cdots&1\\ \vdots&\vdots&\ddots&\vdots\\1&1&\cdots&1 \end{bmatrix}\begin{bmatrix}1\\1\\ \vdots\\1 \end{bmatrix} = \begin{bmatrix}n\\n\\ \vdots\\n \end{bmatrix} = n\begin{bmatrix}1\\1\\ \vdots\\1 \end{bmatrix}

    doesn't that make (1,1,1,...,1) an eigenvector corresponding to the eigenvalue n?
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  10. #25
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    Re: Eigenvalues

    Dealing with the same matrix but the diagonals now all zeros, the characteristic equation would be (n-1-\lambda)(-\lambda-1)^{n-1}
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