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Math Help - eigen values of transpose

  1. #1
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    eigen values of transpose

    A is a square matrix over F field
    if k is the eigen value of A

    prove that k is eigen value of A^t too
    and has the same eigen vectors
    ??

    eigen vectors are the solution space P(A)
    is found by solving (A-kI)x=0
    dim P(A)=dim n -dim (ro(a))
    rho(a)=rho(a^t)
    |A|=|A^t|

    these are the laws i maneged to come up with to solve it

    ??
    Last edited by transgalactic; June 22nd 2011 at 07:07 AM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: eigen values of transpose

    Quote Originally Posted by transgalactic View Post
    A is a square matrix over F field if k is the eigen value of A prove that k is eigen value of A^t too

    \chi(\lambda)=\det (A-\lambda I)=\det ((A-\lambda I)^t)=\det(A^t-\lambda I) so, A and A^t have the same characteristic polynomial.


    and has the same eigen vectors??
    Not necessarily.
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    Re: eigen values of transpose

    thanks
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    Re: eigen values of transpose

    why its not nessesery?
    why if we have the same eigen values then we have the same eigen vector(our matrices differ)
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    Re: eigen values of transpose

    Quote Originally Posted by transgalactic View Post
    why its not nessesery?
    why if we have the same eigen values then we have the same eigen vector(our matrices differ)
    What are you trying to ask??

    Just because two different matrices have an eigenvalue in common doesn't mean that the eigenvector is the same in both cases. Just try it on some small examples to see that the vector can change.

    By the way..... spell check really is a wonderful thing. It really helps us answer your questions if they are written correctly and neatly.
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    Re: eigen values of transpose

    i am asked to prove that their eigen vector are the same too .,for evey eigen value
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    MHF Contributor FernandoRevilla's Avatar
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    Re: eigen values of transpose

    Quote Originally Posted by transgalactic View Post
    i am asked to prove that their eigen vector are the same too .,for evey eigen value

    Choose A=\begin{bmatrix}{0}&{1}\\{0}&{0}\end{bmatrix}

    The eigenvectors of A are different from the eigenvectors of A^t .
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  8. #8
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    Re: eigen values of transpose

    Well, you can't. As you have been told that is not true.

    For example, if A= \begin{bmatrix}5 & -2 \\ 6 & -2\end{bmatrix} has eigenvalues 1 and 2 with corresponding eigenvectors \begin{bmatrix}1 \\ 2\end{bmatrix} and \begin{bmatrix}2 \\ 5\end{bmatrix}, respectively.

    Its transpose is A^*= \begin{bmatrix}5 & 6 \\ -2 & -2\end{bmatrix} which also has eigenvalues 1 and 2 but with corresponding eigenvectors \begin{bmatrix}3\\ -2\end{bmatrix} and \begin{bmatrix}-2 \\ 1\end{bmatrix}, respectively. Those eigenvectors are not multiples of the eigenvectors of A.
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