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Math Help - Eigenvectors

  1. #1
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    Talking Eigenvectors

    I have the matrix

    A=\left[ \begin{matrix} i & 1 \\ 0 & (-1+i) \end{matrix} \right]

    With the eigenvalues

    \lambda_1 = i, \lambda_2 = (-1+i)

    then the eigenvectors

    \lambda_1 : \left[ \begin{matrix} i - i & 1 \\ 0 & (-1+i) - i \end{matrix}\right] \sim \left[\begin{matrix} 0 & 1 \\ 0 & -1 \end{matrix}\right] \sim \left[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right]

    \lambda_2 : \left[ \begin{matrix} i - (-1+i) & 1 \\ 0 & (-1+i) - (-1+i) \end{matrix}\right] \sim \left[\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}\right]

    Now I understand that for \lambda_2 the eigenvector is \left[ 1 , -1\right] since  x_1 + x_2 = 0 \rightarrow x_1 = -x_2.

    However when it comes to the eigenvector for \lambda_1 all I have (in reduced form) x_2 = 0.
    Why is the eigenvector for \lambda_1 \rightarrow \left[ 1 , 0 \right]?

    Is it because the first column (x_1) is a free variable?
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  2. #2
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    Re: Eigenvectors

    because any vector of the form (a,0) = a(1,0) satisfies x2 = 0, so this is an eigenvector iff a ≠ 0.

    yes this is the same as being a "free variable", any choice of x1 (freely made) will do (but of course, the fact that an eigenvector must be non-zero means we can't choose x1 = 0, although (0,0) *is* in the eigenspace Eλ1).
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  3. #3
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    Re: Eigenvectors

    First, do you understand that there is no such thing as "the eigenvector" for given eigenvalue? Given any eigenvector, any (non-zero) multiple of it is also an eigenvector. In fact, the set of all eigenvectors for a given eigenvalue form a subspace. You say that "the eigenvector" for eigenvalue \lambda_2 is [1, -1]. In fact, any vector of the form [a, -a], for a any (non-zero) number is an eigenvector.

    I had to include "(non-zero)" because Deveno says "but of course, the fact that an eigenvector must be non-zero". Some text (but, admittedly a minority) define an eigenvalue by saying " \lambda is an eigenvalue for linear operator A if and only if there exist a non-zero vector Av= \lambda v" but then define an "eigenvector" as any vector satisfying Av= \lambda v. That has the advantage, in my opinion, of including the 0 vector as an "eigenvector" so that the set of all eigenvector corresponding to a given eigenvalue is a subspace without having to add the 0 vector separately.
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