Perturbation theory of the eigenvalues about a symmetric matrix: Reality of Eigenvalues












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Let $A$ be an $ntimes n$ real symmetric matrix, and let $E$ be a real matrix. Is it true that if the perturbation matrix $E$ is small in some norm, then the eigenvalues of $hat A := A+E$ are all real? What if we have more properties on $A$ and $E$: $A$ is diagonalizable and $E$ is skew symmetric?



Note:
I have seen two interesting lines of discussions on the eigenvalues of a symmetric matrix with perturbation here and here, but there is no discussion on the reality of eigenvalues.
Thanks.










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  • 2




    Have you lookes at examples like :$$A= begin{bmatrix} 0 & 0 \ 0 & 0 \ end{bmatrix} $$and $$E= begin{bmatrix} 0 & epsilon \ - epsilon & 0 \ end{bmatrix}. $$
    – Keith McClary
    Jan 4 at 22:35






  • 1




    Going beyond Keith's excellent example, it doesn't matter what $A$ is, there are always going to be matrices arbitrarily close to $A$ with non-real eigenvalues. This is pretty much because the reals have no interior as a subset of the complex numbers, so all real numbers are approached by non-reals.
    – Paul Sinclair
    Jan 5 at 3:50










  • Thank you very much for the excellent example and the excellent intuition!
    – Arthur
    Jan 5 at 5:38
















0














Let $A$ be an $ntimes n$ real symmetric matrix, and let $E$ be a real matrix. Is it true that if the perturbation matrix $E$ is small in some norm, then the eigenvalues of $hat A := A+E$ are all real? What if we have more properties on $A$ and $E$: $A$ is diagonalizable and $E$ is skew symmetric?



Note:
I have seen two interesting lines of discussions on the eigenvalues of a symmetric matrix with perturbation here and here, but there is no discussion on the reality of eigenvalues.
Thanks.










share|cite|improve this question


















  • 2




    Have you lookes at examples like :$$A= begin{bmatrix} 0 & 0 \ 0 & 0 \ end{bmatrix} $$and $$E= begin{bmatrix} 0 & epsilon \ - epsilon & 0 \ end{bmatrix}. $$
    – Keith McClary
    Jan 4 at 22:35






  • 1




    Going beyond Keith's excellent example, it doesn't matter what $A$ is, there are always going to be matrices arbitrarily close to $A$ with non-real eigenvalues. This is pretty much because the reals have no interior as a subset of the complex numbers, so all real numbers are approached by non-reals.
    – Paul Sinclair
    Jan 5 at 3:50










  • Thank you very much for the excellent example and the excellent intuition!
    – Arthur
    Jan 5 at 5:38














0












0








0







Let $A$ be an $ntimes n$ real symmetric matrix, and let $E$ be a real matrix. Is it true that if the perturbation matrix $E$ is small in some norm, then the eigenvalues of $hat A := A+E$ are all real? What if we have more properties on $A$ and $E$: $A$ is diagonalizable and $E$ is skew symmetric?



Note:
I have seen two interesting lines of discussions on the eigenvalues of a symmetric matrix with perturbation here and here, but there is no discussion on the reality of eigenvalues.
Thanks.










share|cite|improve this question













Let $A$ be an $ntimes n$ real symmetric matrix, and let $E$ be a real matrix. Is it true that if the perturbation matrix $E$ is small in some norm, then the eigenvalues of $hat A := A+E$ are all real? What if we have more properties on $A$ and $E$: $A$ is diagonalizable and $E$ is skew symmetric?



Note:
I have seen two interesting lines of discussions on the eigenvalues of a symmetric matrix with perturbation here and here, but there is no discussion on the reality of eigenvalues.
Thanks.







linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition perturbation-theory






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share|cite|improve this question











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asked Jan 4 at 17:12









ArthurArthur

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47112








  • 2




    Have you lookes at examples like :$$A= begin{bmatrix} 0 & 0 \ 0 & 0 \ end{bmatrix} $$and $$E= begin{bmatrix} 0 & epsilon \ - epsilon & 0 \ end{bmatrix}. $$
    – Keith McClary
    Jan 4 at 22:35






  • 1




    Going beyond Keith's excellent example, it doesn't matter what $A$ is, there are always going to be matrices arbitrarily close to $A$ with non-real eigenvalues. This is pretty much because the reals have no interior as a subset of the complex numbers, so all real numbers are approached by non-reals.
    – Paul Sinclair
    Jan 5 at 3:50










  • Thank you very much for the excellent example and the excellent intuition!
    – Arthur
    Jan 5 at 5:38














  • 2




    Have you lookes at examples like :$$A= begin{bmatrix} 0 & 0 \ 0 & 0 \ end{bmatrix} $$and $$E= begin{bmatrix} 0 & epsilon \ - epsilon & 0 \ end{bmatrix}. $$
    – Keith McClary
    Jan 4 at 22:35






  • 1




    Going beyond Keith's excellent example, it doesn't matter what $A$ is, there are always going to be matrices arbitrarily close to $A$ with non-real eigenvalues. This is pretty much because the reals have no interior as a subset of the complex numbers, so all real numbers are approached by non-reals.
    – Paul Sinclair
    Jan 5 at 3:50










  • Thank you very much for the excellent example and the excellent intuition!
    – Arthur
    Jan 5 at 5:38








2




2




Have you lookes at examples like :$$A= begin{bmatrix} 0 & 0 \ 0 & 0 \ end{bmatrix} $$and $$E= begin{bmatrix} 0 & epsilon \ - epsilon & 0 \ end{bmatrix}. $$
– Keith McClary
Jan 4 at 22:35




Have you lookes at examples like :$$A= begin{bmatrix} 0 & 0 \ 0 & 0 \ end{bmatrix} $$and $$E= begin{bmatrix} 0 & epsilon \ - epsilon & 0 \ end{bmatrix}. $$
– Keith McClary
Jan 4 at 22:35




1




1




Going beyond Keith's excellent example, it doesn't matter what $A$ is, there are always going to be matrices arbitrarily close to $A$ with non-real eigenvalues. This is pretty much because the reals have no interior as a subset of the complex numbers, so all real numbers are approached by non-reals.
– Paul Sinclair
Jan 5 at 3:50




Going beyond Keith's excellent example, it doesn't matter what $A$ is, there are always going to be matrices arbitrarily close to $A$ with non-real eigenvalues. This is pretty much because the reals have no interior as a subset of the complex numbers, so all real numbers are approached by non-reals.
– Paul Sinclair
Jan 5 at 3:50












Thank you very much for the excellent example and the excellent intuition!
– Arthur
Jan 5 at 5:38




Thank you very much for the excellent example and the excellent intuition!
– Arthur
Jan 5 at 5:38










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