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Math Help - Left Eigenvectors

  1. #1
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    Left Eigenvectors

    I have two problems dealing with left eigenvectors:

    (1) Show that a left eigenvector y correcpinding to the eigenvalue of A belonging to M_n is a right eigenvector of A* corresponding to (with a bar over it).

    (2) Show that if y is a left eigenvector corresponding to the eigenvalue of A belonging to M_n then y(with a bar over it) is a right eigenvector of A^T corresponding to .

    I know that the definition of left eigenvector states...the vector y ≠ 0 is said to be a left eigenvector of A if it satisfies y* A = y*. Then is called a left eigenvector of A.
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  2. #2
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    Re: Left Eigenvectors

    if y is a left eigenvector of A, then y*A = λy*. therefore, (y*A)* = (λy*)*, that is:

    A*y = λ*y.

    (writing λ* for the complex conjugate of λ).

    the trick is noticing that if:

    y^* = \begin{bmatrix}\overline{y_1}\\ \overline{y_2}\\ \vdots\\ \overline{y_n}\end{bmatrix} then:

    \lambda y^* = \begin{bmatrix}\lambda \overline{y_1}\\ \lambda \overline{y_2}\\ \vdots\\ \lambda \overline{y_n}\end{bmatrix} so:

    (\lambda y^*)^* = \begin{bmatrix} \overline{\lambda} y_1, \overline{\lambda} y_2, \dots, \overline{\lambda} y_n \end{bmatrix} = \overline{\lambda}y

    for the second question, note that y*A = λy*, means that y*(A - λI) = 0^t (the 0 row-vector), so

    (y^*(A - \lambda I))^T = 0 that is: (A - \lambda I)^T\overline{y} = 0 so that:

    A^T\overline{y} = \lambda\overline{y}
    Last edited by Deveno; November 27th 2011 at 07:55 AM.
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  3. #3
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    Re: Left Eigenvectors

    Must the left eigenvectors and right eigenvectors of a matrix be the same? Prove or give counter example.

    I want to say no, but am not sure if I am even correct let alone proving. Can someone help me out? Thanks!
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