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Math Help - Eigenvalues & eigenvectors of an n*n matrix

  1. #1
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    Eigenvalues & eigenvectors of an n*n matrix

    Hello,

    I need some help with this problem:
    \begin{pmatrix} 1 & 0 & 1 & 0 & \cdots \\ 0 & 1 & 0 & 1 & \cdots \\ 1 & 0 & 1 & 0 & \cdots \\ 0 & 1 & 0 & 1 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}

    The instructions are to find the eigenvalues and eigenvectors of this n by n matrix. I've tried to use det(A -\lambda I_n) = 0 to get the characteristic polynomial, but I can't seem to finish it.

    I'd be grateful for any hints.
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    Re: Eigenvalues & eigenvectors of an n*n matrix

    Here are a few observations that may be useful.

    First, The \text{Rank}(A)=\text{Rank}(A-0\cdot I)=2 for all n.

    This gives that \text{Null}(A)=n-2 so zero is an eigenvalue of multiplicity n-2.

    When you compute A-0\cdot I=A and reduce to row echelon form you can see a pattern for the eigenvectors. Just write out a few small cases and look for the pattern.

    For the case when n is even we get

    A^2=nA \iff A^2-\frac{n}{2}A=0 This implies that the polynomial p(x)=x^2-\frac{n}{2}x=x(x-\frac{n}{2}) is in the annihilating Ideal, so the minimum polynomial must divide p(x)

    so the only possible eigenvalues are 0 and \frac{n}{2}. The two cases I checked admitted 2 eigenvectors for the eigenvalue \frac{n}{2}

    When I computed the the matrix A-\frac{n}{2}I and reduced to row echelon form for n=4 and n=6 there seems to be a pattern for the eigenvectors . I hope this helps.

    I haven't discovered anything useful for the case when n is odd.
    Last edited by TheEmptySet; April 12th 2012 at 12:48 PM.
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    Re: Eigenvalues & eigenvectors of an n*n matrix

    Thanks for help, I managed to finish it
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