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Thread: eigenvalues of AB and of BA

  1. #1
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    eigenvalues of AB and of BA

    Is it true that the eigenvalues of AB are the same as the eigenvalues of BA?
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    Quote Originally Posted by math8 View Post
    Is it true that the eigenvalues of AB are the same as the eigenvalues of BA?
    No, multiplication of matrices isn't commutative.
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    I know AB is not BA, but I read somewhere that their eigenvalues (trace and determinant also) are the same. I just wanted to get a confirmation of this.
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    Quote Originally Posted by math8 View Post
    I know AB is not BA, but I read somewhere that their eigenvalues (trace and determinant also) are the same. I just wanted to get a confirmation of this.
    Sorry for the mistake at first.

    $\displaystyle det(AB-\lambda I)=det(AB-\lambda AA^{-1})=det(A(B-\lambda A^{-1}))=det(A)det(B-\lambda A^{-1})$
    $\displaystyle =det(B-\lambda A^{-1})det(A)=det((B-\lambda A^{-1})A)=det(BA-\lambda I)$

    I am pretty sure I have proving it here as well as some other proofs that may help in Linear:
    http://www.mathhelpforum.com/math-he...rexamples.html
    Last edited by dwsmith; May 14th 2010 at 05:40 PM. Reason: Wasn't thinking
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  5. #5
    Super Member Random Variable's Avatar
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    It's easy to prove.


    Let $\displaystyle \lambda_{1} $ be an eigenvalue of $\displaystyle AB$

    then $\displaystyle ABx=\lambda_{1} x$

    $\displaystyle BABx = \lambda_{1} B x $

    which means that $\displaystyle \lambda_{1} $ is an eigenvalue of $\displaystyle BA $ with associated eigenvector $\displaystyle Bx $


    Now let $\displaystyle \lambda_{2} $ be an eigenvalue of $\displaystyle BA $

    then $\displaystyle BAx=\lambda_{2}x $

    $\displaystyle ABAx = \lambda_{2}Ax$

    which means that $\displaystyle \lambda_{2} $ is an eigenvalue of $\displaystyle AB $ with associated eigenvector $\displaystyle Ax$

    QED
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  6. #6
    Senior Member roninpro's Avatar
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    Quote Originally Posted by dwsmith View Post
    Sorry for the mistake at first.

    $\displaystyle det(AB-\lambda I)=det(AB-\lambda AA^{-1})=det(A(B-\lambda A^{-1}))=det(A)det(B-\lambda A^{-1})$
    $\displaystyle =det(B-\lambda A^{-1})det(A)=det((B-\lambda A^{-1})A)=det(BA-\lambda I)$

    I am pretty sure I have proving it here as well as some other proofs that may help in Linear:
    http://www.mathhelpforum.com/math-he...rexamples.html
    There might be a slight hiccup. You don't know whether or not $\displaystyle A$ or $\displaystyle B$ is invertible.
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    Fortunately, Random Variables' proof, in addition to being simpler, does not require that A and B be invertible.
    Last edited by HallsofIvy; May 15th 2010 at 02:50 PM.
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    Quote Originally Posted by Random Variable View Post
    It's easy to prove.


    Let $\displaystyle \lambda_{1} $ be an eigenvalue of $\displaystyle AB$

    then $\displaystyle ABx=\lambda_{1} x$

    $\displaystyle BABx = \lambda_{1} B x $

    which means that $\displaystyle \lambda_{1} $ is an eigenvalue of $\displaystyle BA $ with associated eigenvector $\displaystyle Bx $
    $\displaystyle BABx=\lambda Bx$ doesn't necessarily imply that $\displaystyle \lambda$ is an eigenvalue of $\displaystyle BA$ because we might have $\displaystyle Bx=0.$
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    Super Member Random Variable's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    $\displaystyle BABx=\lambda Bx$ doesn't necessarily imply that $\displaystyle \lambda$ is an eigenvalue of $\displaystyle BA$ because we might have $\displaystyle Bx=0.$
    Deal with the case of $\displaystyle A$ or $\displaystyle B$ being zero matrices separately.


    If $\displaystyle A $ or $\displaystyle B$ are zero matrices, then $\displaystyle AB=BA=0$, and the only eigenvalue of a zero matrix is $\displaystyle \lambda = 0 $
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    Quote Originally Posted by Random Variable View Post
    Deal with the case of $\displaystyle A$ or $\displaystyle B$ being zero matrices separately.


    If $\displaystyle A $ or $\displaystyle B$ are zero matrices, then $\displaystyle AB=BA=0$, and the only eigenvalue of a zero matrix is $\displaystyle \lambda = 0 $
    That's not NonCommAlg's point. You must consider the possiblity that neither A nor B are zero matrices, that $\displaystyle ABx= \lambda x$ for some non-zero vector x, but Bx= 0. Of course, that's also easy to answer- if Bx= 0, then ABx= 0 also so the eigenvalue for both AB and BA is 0.
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  11. #11
    Super Member Random Variable's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    That's not NonCommAlg's point. You must consider the possiblity that neither A nor B are zero matrices, that $\displaystyle ABx= \lambda x$ for some non-zero vector x, but Bx= 0. Of course, that's also easy to answer- if Bx= 0, then ABx= 0 also so the eigenvalue for both AB and BA is 0.
    Yeah, I forgot about that possibility.
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  12. #12
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    I know I have this proof covering all cases, and once I find it, I will post it.
    I am not sure if the red is necessarily correct.
    AB and BA have the same eigenvalues iff. AB is similar to BA.

    AB=C
    BA=D

    If C and D are similar matrices, then there exist S such that $\displaystyle C=S^{-1}DS$.

    $\displaystyle p_c(\lambda)=det(AB-\lambda I)=det(C-\lambda I)=det(S^{-1}DS-\lambda I)$
    $\displaystyle =det(S^{-1}DS-\lambda S^{-1}IS)=det(S^{-1}(D-\lambda I)S)$
    $\displaystyle =det(S^{-1}S)det(D-\lambda I)=det(BA-\lambda I)=p_d(\lambda)$
    Last edited by dwsmith; May 17th 2010 at 11:03 AM.
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