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Math Help - [SOLVED] Proof question for eigenvectors

  1. #1
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    [SOLVED] Proof question for eigenvectors

    The vectors v_1, v_2, ..., v_k in the vector space R^n are eigenvectors for the matrix A corresponding to the eigenvalues \lambda_1, \lambda_2, ..., \lambda_k (not necessarily distinct).

    If these vectors are linearly independent and \lambda_{k+1} is distinct from all of the other \lambda's, prove that \{ v_1, v_2, ..., v_k, v_{k+1} \} is linearly independent for any eigenvector v_{k+1} corresponding to \lambda_{k+1}.

    (Hint: if not, we have v_{k+1}=c_1v_1+c_2v_2+...+c_kv_k for some constants and Av_{k+1} can easily be written in two ways.)

    Please show your work.
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  2. #2
    Member Mauritzvdworm's Avatar
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    suppose that \{v_1,\dots,v_k,v_{k+1}\} is not linearly independant, hence we can write

    v_{k+1}=c_1 v_1+\dots +c_k v_k
    now consider the following

    Av_{k+1}=\lambda_{k+1}v_{k+1}=\lambda_{k+1}\left(c  _1 v_1+\dots +c_k v_k\right)

    and we also have

    Av_{k+1}=A\left(c_1 v_1+\dots +c_k v_k\right)=\lambda_1 c_1 v_1+\dots +\lambda_k c_k v_k
    clearly Av_{k+1}-Av_{k+1}=0

    hence

    \left(\lambda_{k+1} -\lambda_1\right)c_1 v_1+\dots+\left(\lambda_{k+1} -\lambda_k\right)c_k v_k=0

    and since \{v_1,\dots,v_k\} is linearly independant we know this equality can only hold when the constants are zero, hence

    \lambda_{k+1}=\lambda_{i} for each i=1,2,\dots,k

    so \lambda_{k+1} is not distinct.

    So this concludes the proof with a contrapositive argument.
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