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Thread: Inverse Matrix - Eigenvalues & Eigenvectors (Proof Needed)

  1. #1
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    Inverse Matrix - Eigenvalues & Eigenvectors (Proof Needed)

    Hello,

    One of the questions I was presented in a tutorial the other day really stumped me and I am unsure as to how to prove it.
    The question is:

    "Suppose that A is an invertible matrix and that x is an eigenvector for A with eigenvalue 'lambda cannot equal 0'. Show that x is an eigenvector for the inverse matrix of A with eigenvalue 'lambda-1'.

    I have been shown a similar question where you had to prove that the matrix A2 had an eigenvalue of lambda2 through the manipulation of the equation Ax = (lambda)x as below:

    Ax = (lambda)x
    A(Ax) = A(lambda)x
    A2x = (lambda)Ax
    A2x = (lambda)(lambda)x
    A2x = (lambda)2x

    Is the solution to this question reached through the same steps, or is there a step I need to do differently?

    Any help would be greatly appreciated
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  2. #2
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    Re: Inverse Matrix - Eigenvalues & Eigenvectors (Proof Needed)

    Perhaps you should attempt this question before asking for help. Don't be afraid to try.

    $\displaystyle Ax=\lambda x \iff A^{-1}Ax=A^{-1}\lambda x \iff x=\lambda A^{-1}x$
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  3. #3
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    Re: Inverse Matrix - Eigenvalues & Eigenvectors (Proof Needed)

    Let $\displaystyle A $ be a matrix with eigenvalue $\displaystyle \lambda $ and eigenvector $\displaystyle x $.

    Then $\displaystyle Ax = \lambda x$

    Since $\displaystyle A $ is invertible, we can multiply both sides by $\displaystyle A^{-1} $.

    $\displaystyle A^{-1} Ax = A^{-1}\lambda x $
    $\displaystyle x = A^{-1} \lambda x $

    Solve for $\displaystyle A^{-1} x$.

    Thus, $\displaystyle A^{-1} x = \frac{1}{\lambda} x$
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