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Thread: different eigenvalues

  1. #1
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    different eigenvalues

    Suppose you find n different eigenvalues for an nxn matrix. Are you then guaranteed, that the eigenvectors corresponding to those eigenvalues are linearly independent?
    If so, from what does that follow?
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: different eigenvalues

    Yes. Suppose you have n different eigenvalues $\displaystyle c_1, \ldots, c_n$ of an endomorfism $\displaystyle f: V\to V$ corresponding to the eigenvectors $\displaystyle v_1, \ldots, v_n$ then the set $\displaystyle \{v_1, \ldots, v_n\}$ is linearly independent.

    Suppose the eigenvectors are linearly dependent, then we know there's a first vector $\displaystyle v_k$ in the row which can be written as a linear combination of the previous vectors $\displaystyle v_1, \ldots, v_{k-1}$. Thus $\displaystyle \forall a_i \in K$:
    $\displaystyle v_k=a_1v_1+\ldots+a_{k-1}v_{k-1}$
    and $\displaystyle v_1, \ldots, v_{k-1}$ are linearly independent.
    We have:
    $\displaystyle f(v_k)=f(a_1v_1+\ldots+a_{k-1}v_{k-1})=a_1c_1v_1+\ldots+a_{k-1}c_{k-1}v_{k-1}$ and
    $\displaystyle f(v_k)=c_kv_k=c_ka_1v_1+\ldots+c_ka_{k-1}v_{k-1}$

    Thus:
    $\displaystyle a_1c_1v_1+\ldots+a_{k-1}c_{k-1}v_{k-1}=c_ka_1v_1+\ldots+c_ka_{k-1}v_{k-1}$
    $\displaystyle \Rightarrow (c_1-c_k)a_1v_1+\ldots+(c_{k-1}-c_k)a_{k-1}v_{k-1}=0$

    But because $\displaystyle v_1,\ldots,v_{k-1}$ are linearly independent alle the coefficients are equal to zero, so:
    $\displaystyle (c_1-c_k)a_1=\ldots=(c_{k-1}-c_k)a_{k-1}=0$
    But because all the eigenvalues are different $\displaystyle a_1=\ldots=a_{k-1}=0$ and so $\displaystyle v_k=0$, this is a contradiction because $\displaystyle v_k$ wa supposed to be an eigenvector so our statement is wrong and the eigenvectors $\displaystyle v_1, \ldots,v_k$ are indeed linearly independent.
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