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.
linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition perturbation-theory
asked Jan 4 at 17:12
ArthurArthur
52912
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add a comment |
$begingroup$
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
asked Jan 4 at 17:12
ArthurArthur
52912
$endgroup$
2
$begingroup$
Have you lookes at examples like :$$A= begin{bmatrix} 0 & 0 \ 0 & 0 \ end{bmatrix} $$and $$E= begin{bmatrix} 0 & epsilon \ - epsilon & 0 \ end{bmatrix}. $$
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– Keith McClary
Jan 4 at 22:35
1
$begingroup$
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.
$endgroup$
– Paul Sinclair
Jan 5 at 3:50
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Thank you very much for the excellent example and the excellent intuition!
$endgroup$
– Arthur
Jan 5 at 5:38
add a comment |
$begingroup$
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
asked Jan 4 at 17:12
ArthurArthur
52912
$endgroup$
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
linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition perturbation-theory
asked Jan 4 at 17:12
ArthurArthur
52912
asked Jan 4 at 17:12
ArthurArthur
52912
asked Jan 4 at 17:12
ArthurArthur
52912
asked Jan 4 at 17:12
ArthurArthur
52912
asked Jan 4 at 17:12
ArthurArthur
52912
52912
2
$begingroup$
Have you lookes at examples like :$$A= begin{bmatrix} 0 & 0 \ 0 & 0 \ end{bmatrix} $$and $$E= begin{bmatrix} 0 & epsilon \ - epsilon & 0 \ end{bmatrix}. $$
$endgroup$
– Keith McClary
Jan 4 at 22:35
1
$begingroup$
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.
$endgroup$
– Paul Sinclair
Jan 5 at 3:50
$begingroup$
Thank you very much for the excellent example and the excellent intuition!
$endgroup$
– Arthur
Jan 5 at 5:38
add a comment |
2
$begingroup$
Have you lookes at examples like :$$A= begin{bmatrix} 0 & 0 \ 0 & 0 \ end{bmatrix} $$and $$E= begin{bmatrix} 0 & epsilon \ - epsilon & 0 \ end{bmatrix}. $$
$endgroup$
– Keith McClary
Jan 4 at 22:35
1
$begingroup$
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.
$endgroup$
– Paul Sinclair
Jan 5 at 3:50
$begingroup$
Thank you very much for the excellent example and the excellent intuition!
$endgroup$
– Arthur
Jan 5 at 5:38
2
2
$begingroup$
Have you lookes at examples like :$$A= begin{bmatrix} 0 & 0 \ 0 & 0 \ end{bmatrix} $$and $$E= begin{bmatrix} 0 & epsilon \ - epsilon & 0 \ end{bmatrix}. $$
$endgroup$
– Keith McClary
Jan 4 at 22:35
$begingroup$
Have you lookes at examples like :$$A= begin{bmatrix} 0 & 0 \ 0 & 0 \ end{bmatrix} $$and $$E= begin{bmatrix} 0 & epsilon \ - epsilon & 0 \ end{bmatrix}. $$
$endgroup$
– Keith McClary
Jan 4 at 22:35
1
1
$begingroup$
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.
$endgroup$
– Paul Sinclair
Jan 5 at 3:50
$begingroup$
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.
$endgroup$
– Paul Sinclair
Jan 5 at 3:50
$begingroup$
Thank you very much for the excellent example and the excellent intuition!
$endgroup$
– Arthur
Jan 5 at 5:38
$begingroup$
Thank you very much for the excellent example and the excellent intuition!
$endgroup$
– Arthur
Jan 5 at 5:38
add a comment |
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$begingroup$
Have you lookes at examples like :$$A= begin{bmatrix} 0 & 0 \ 0 & 0 \ end{bmatrix} $$and $$E= begin{bmatrix} 0 & epsilon \ - epsilon & 0 \ end{bmatrix}. $$
$endgroup$
– Keith McClary
Jan 4 at 22:35
1
$begingroup$
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.
$endgroup$
– Paul Sinclair
Jan 5 at 3:50
$begingroup$
Thank you very much for the excellent example and the excellent intuition!
$endgroup$
– Arthur
Jan 5 at 5:38