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Thread: 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:
    $\displaystyle \begin{pmatrix} 1 & 0 & 1 & 0 & \cdots \\ 0 & 1 & 0 & 1 & \cdots \\ 1 & 0 & 1 & 0 & \cdots \\ 0 & 1 & 0 & 1 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}$

    The instructions are to find the eigenvalues and eigenvectors of this n by n matrix. I've tried to use $\displaystyle det(A -\lambda I_n) = 0$ 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 $\displaystyle \text{Rank}(A)=\text{Rank}(A-0\cdot I)=2$ for all n.

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

    When you compute $\displaystyle A-0\cdot I=A$ 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 $\displaystyle n$ is even we get

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

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

    When I computed the the matrix $\displaystyle A-\frac{n}{2}I$ 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; Apr 12th 2012 at 11:48 AM.
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    Re: Eigenvalues & eigenvectors of an n*n matrix

    Thanks for help, I managed to finish it
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