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Math Help - Eigenevectors and Eigenvalues

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    Eigenevectors and Eigenvalues

    Suppose A is a symmetric N N matrix with eigenvectors vi, i = 1, 2, 3 N with
    corresponding eigenvalues λi, i = 1, 2,3 N.

    Pick any two distinct eigenvalues (assuming such a pair exists). Let's call them λ1 and λ 2 and their corresponding eigenvectors v1 and v2.

    (a) Write down the matrix equations that show that v1 and v2 are eigenvectors of A.

    (b) Compute the transpose of the equation satisfied by v2.

    (c) Multiply, from the right, the result of part (b) by v1.

    (d) Use the assumptions that A is symmetric and λ1≠ λ 2 to deduce a value for v2^t v1.

    (e) What important property can you deduce from your result in part (d)?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Eigenevectors and Eigenvalues

    Quote Originally Posted by Ants View Post
    Suppose A is a symmetric N N matrix with eigenvectors vi, i = 1, 2, 3 N with
    corresponding eigenvalues λi, i = 1, 2,3 N.

    Pick any two distinct eigenvalues (assuming such a pair exists). Let's call them λ1 and λ 2 and their corresponding eigenvectors v1 and v2.

    (a) Write down the matrix equations that show that v1 and v2 are eigenvectors of A.
    Av_1=\lambda_1v_1,\;Av_2=\lambda_2v_2

    (b) Compute the transpose of the equation satisfied by v2.
    Av_2=\lambda_2v_2\Rightarrow v_2^tA^t=\lambda_2v_2^t\Rightarrow (A\mbox{ symmetric})\;v_2^tA=\lambda_2v_2^t

    (c) Multiply, from the right, the result of part (b) by v1.
    v_2^tAv_1=\lambda_2v_2^tv_1

    (d) Use the assumptions that A is symmetric and λ1≠ λ 2 to deduce a value for v2^t v1.
    From (a), v_2^tAv_1=\lambda_1v_2^tv_1 . Using \lambda_1\neq \lambda_2 :

    \begin{Bmatrix} v_2^tAv_1=\lambda_1v_2^tv_1\\ v_2^tAv_1=\lambda_2v_2^tv_1 \end{matrix}\Rightarrow 0=(\lambda_1-\lambda_2)v_2^tv_1\Rightarrow v_2^tv_1=0

    (e) What important property can you deduce from your result in part (d)?
    v_2^tv_1=<v_2,v_1>. (Conclude).

    P.D. Please, next time show some work (according to the rules of MHF).
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  3. #3
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    Re: Eigenevectors and Eigenvalues

    Hey Ants.

    I'll start by mentioning that if Ax = lx, then x is an eigenvector and l is the corresponding eigen-value and this is the standard definition.
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