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Thread: Eigenvalue Questions

  1. #1
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    Eigenvalue Questions



    I can do i) but just a bit stuck on ii).

    my working for i) is: E^2-E=0 so that means $\displaystyle \lambda^2-\lambda$ is an eigenvalue of E^2-E=0, since the zero vector only has an eigenvalue of 0, then $\displaystyle \lambda^2-\lambda=0$ and so \lambda = 0 or 1

    cheers

    [hr]



    I sorta have an idea but dono how to finish this Q

    If we let x be a eigenvector of $\displaystyle B^TB$ corresponding to $\displaystyle \lambda$, then we have $\displaystyle B^TBx = \lambda x$

    Now just playing around: $\displaystyle B^TBx.x = x. ((B^TB)^Tx)=x.(B^TB)x= x.(\lambda x)$

    but dono how to get to the result $\displaystyle \lambda \ge 0$
    Last edited by Ackbeet; Jun 20th 2011 at 05:00 AM.
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  2. #2
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    Re: Eigenvalue Questions

    for (ii): from part (i) we know that the eigenvalues for E are all 1 (since none are 0). in particular, the eigenspace E0 contains only the 0-vector

    (there are NO non-zero vectors v for which Ev = 0). that is, ker(E) = {0}, or nullity(E) = 0.

    thus dim(V) = rank(E), so E is invertible (it is 1-1 and onto). hence from E^2 = E we have:

    E^2(E^-1) = E(E^-1)

    E(EE^-1) = I........so ?

    for question 2: note that $\displaystyle x^T(B^TB)x = <Bx,Bx>\ \ge 0$

    note as well that $\displaystyle x^T(B^TB)x = x^T(\lambda x) = \lambda(x^Tx) = \lambda<x,x>$

    can <x,x> be negative?
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: Eigenvalue Questions

    An alternative for the first question: $\displaystyle E^2=E\Rightarrow E(Ex)=Ex$ for all $\displaystyle x\in\mathbb{R}^n$. Consider $\displaystyle x\neq 0$ , if $\displaystyle Ex=0$ then $\displaystyle x$ is an eigenvector associated to $\displaystyle \lambda=0$, if $\displaystyle Ex\neq 0$ then $\displaystyle Ex$ is an eigenvector associated to $\displaystyle \lambda=1$ .
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