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Thread: [SOLVED] Proof question for eigenvectors

  1. #1
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    [SOLVED] Proof question for eigenvectors

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

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

    (Hint: if not, we have $\displaystyle v_{k+1}=c_1v_1+c_2v_2+...+c_kv_k$ for some constants and $\displaystyle Av_{k+1}$ 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 $\displaystyle \{v_1,\dots,v_k,v_{k+1}\}$ is not linearly independant, hence we can write

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

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

    and we also have

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

    hence

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

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

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

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

    So this concludes the proof with a contrapositive argument.
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