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Thread: eigenspace help

  1. #1
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    eigenspace help

    A is the matrix $\displaystyle \begin{bmatrix} 2 & 2 & -2 \\ 2 & -1 & 4\\ -2 & 4 & -1 \end{bmatrix}$. You are given that A has eigenvalues -6 and 3, with algebaric multiplicities 1 and 2 respectively, and that the eigenspace $\displaystyle V_{3} = span{(2,-1,-2),(2,2,1)} $.

    a) use the given eigenvalues to find the characteristic equation of A. (do not expand det(A-ILAMBDA), but the construct the equation from its roots.
    Hence use the hamilton theorem to find A^{-1} .

    I dont know where to being. Any help appreciated,


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  2. #2
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    Re: eigenspace help

    Hey Tweety.

    Recall that the eigenvalues are the solutions to the characteristic equation which means they are the roots. Recall that if a polynomial has solutions a,b,c then we can factorize that polynomial as (x-a)(x-b)(x-c) = 0.
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  3. #3
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    Re: eigenspace help

    so the characteristic equation is $\displaystyle (\lambda+6)(\lambda-3) $ ?
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  4. #4
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    Re: eigenspace help

    You need to factor in the multiplicities of the roots. (Hint: You will have a squared term).
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  5. #5
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    Re: eigenspace help

    Thank you, but I am not sure how to factor in the multiplicties of the roots?

    is it $\displaystyle 1(\lambda+6)2(\lambda-3) $ ?

    is the equation $\displaystyle 2\lambda^{2}+6\lambda-36 $
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  6. #6
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    Re: eigenspace help

    Basically its (lambda + 6)(lambda - 3)(lambda - 3) = characteristic polynomial = f(lambda)
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