Mixing
Gershgorin circle and complex eigenvalues
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As I understood all complex eigenvalue are coming in complex conjugate pairs. Additionally if all the circle don't overlap so in each circle only one eigenvalue exist and if all the circle's centres are on real axis so not way that complex eigenvalue exit. Am I right?
In addition I would like to know if their is any other information that I can learn about existence of complex eigenvalue from Gershgorin circle?
linear-algebra matrices eigenvalues-eigenvectors gershgorin-sets
asked Jan 30 at 7:27
ChaosPredictorChaosPredictor
1295
$endgroup$
add a comment |
$begingroup$
As I understood all complex eigenvalue are coming in complex conjugate pairs. Additionally if all the circle don't overlap so in each circle only one eigenvalue exist and if all the circle's centres are on real axis so not way that complex eigenvalue exit. Am I right?
In addition I would like to know if their is any other information that I can learn about existence of complex eigenvalue from Gershgorin circle?
linear-algebra matrices eigenvalues-eigenvectors gershgorin-sets
asked Jan 30 at 7:27
ChaosPredictorChaosPredictor
1295
$endgroup$
$begingroup$
If the matrix has complex entries, the eigenvalues can be complex, and not necessarily in conjugate pairs. For a complex matrix, if a Gershgorin circle or a union of more of them do not touch the x-axis, the existence of complex eigenvalue(s) is proven.
$endgroup$
– user376343
Jan 31 at 13:08
add a comment |
$begingroup$
As I understood all complex eigenvalue are coming in complex conjugate pairs. Additionally if all the circle don't overlap so in each circle only one eigenvalue exist and if all the circle's centres are on real axis so not way that complex eigenvalue exit. Am I right?
In addition I would like to know if their is any other information that I can learn about existence of complex eigenvalue from Gershgorin circle?
linear-algebra matrices eigenvalues-eigenvectors gershgorin-sets
asked Jan 30 at 7:27
ChaosPredictorChaosPredictor
1295
$endgroup$
As I understood all complex eigenvalue are coming in complex conjugate pairs. Additionally if all the circle don't overlap so in each circle only one eigenvalue exist and if all the circle's centres are on real axis so not way that complex eigenvalue exit. Am I right?
In addition I would like to know if their is any other information that I can learn about existence of complex eigenvalue from Gershgorin circle?
linear-algebra matrices eigenvalues-eigenvectors gershgorin-sets
linear-algebra matrices eigenvalues-eigenvectors gershgorin-sets
asked Jan 30 at 7:27
ChaosPredictorChaosPredictor
1295
asked Jan 30 at 7:27
ChaosPredictorChaosPredictor
1295
asked Jan 30 at 7:27
ChaosPredictorChaosPredictor
1295
asked Jan 30 at 7:27
ChaosPredictorChaosPredictor
1295
asked Jan 30 at 7:27
ChaosPredictorChaosPredictor
1295
1295
$begingroup$
If the matrix has complex entries, the eigenvalues can be complex, and not necessarily in conjugate pairs. For a complex matrix, if a Gershgorin circle or a union of more of them do not touch the x-axis, the existence of complex eigenvalue(s) is proven.
$endgroup$
– user376343
Jan 31 at 13:08
add a comment |
$begingroup$
If the matrix has complex entries, the eigenvalues can be complex, and not necessarily in conjugate pairs. For a complex matrix, if a Gershgorin circle or a union of more of them do not touch the x-axis, the existence of complex eigenvalue(s) is proven.
$endgroup$
– user376343
Jan 31 at 13:08
$begingroup$
If the matrix has complex entries, the eigenvalues can be complex, and not necessarily in conjugate pairs. For a complex matrix, if a Gershgorin circle or a union of more of them do not touch the x-axis, the existence of complex eigenvalue(s) is proven.
$endgroup$
– user376343
Jan 31 at 13:08
$begingroup$
If the matrix has complex entries, the eigenvalues can be complex, and not necessarily in conjugate pairs. For a complex matrix, if a Gershgorin circle or a union of more of them do not touch the x-axis, the existence of complex eigenvalue(s) is proven.
$endgroup$
– user376343
Jan 31 at 13:08
add a comment |
1 Answer
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If the matrix has real entries, or at least real entries in the main diagonal, all Gershgorin circles intersect the real line. Therefore, the circle theorem cannot establish the existence of complex eigenvalues.
answered Jan 30 at 8:01
PierreCarrePierreCarre
1,692212
$endgroup$
$begingroup$
I think you are wrong, what about this example [0 -1 -1; 1 0 -1; 1 1 0] it has +/-i eigenvalues.
$endgroup$
– ChaosPredictor
Jan 30 at 17:41
$begingroup$
The matrix you mention has complex eigenvalues, sure, but what I say is that the theorem does not help to establish that fact. In the conditions I mentioned, the circles intersect the real line and therefore the theorem never rules out real eigenvalues.
$endgroup$
– PierreCarre
Jan 30 at 17:51
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If the matrix has real entries, or at least real entries in the main diagonal, all Gershgorin circles intersect the real line. Therefore, the circle theorem cannot establish the existence of complex eigenvalues.
answered Jan 30 at 8:01
PierreCarrePierreCarre
1,692212
$endgroup$
$begingroup$
I think you are wrong, what about this example [0 -1 -1; 1 0 -1; 1 1 0] it has +/-i eigenvalues.
$endgroup$
– ChaosPredictor
Jan 30 at 17:41
$begingroup$
The matrix you mention has complex eigenvalues, sure, but what I say is that the theorem does not help to establish that fact. In the conditions I mentioned, the circles intersect the real line and therefore the theorem never rules out real eigenvalues.
$endgroup$
– PierreCarre
Jan 30 at 17:51
add a comment |
$begingroup$
If the matrix has real entries, or at least real entries in the main diagonal, all Gershgorin circles intersect the real line. Therefore, the circle theorem cannot establish the existence of complex eigenvalues.
answered Jan 30 at 8:01
PierreCarrePierreCarre
1,692212
$endgroup$
$begingroup$
I think you are wrong, what about this example [0 -1 -1; 1 0 -1; 1 1 0] it has +/-i eigenvalues.
$endgroup$
– ChaosPredictor
Jan 30 at 17:41
$begingroup$
The matrix you mention has complex eigenvalues, sure, but what I say is that the theorem does not help to establish that fact. In the conditions I mentioned, the circles intersect the real line and therefore the theorem never rules out real eigenvalues.
$endgroup$
– PierreCarre
Jan 30 at 17:51
add a comment |
$begingroup$
If the matrix has real entries, or at least real entries in the main diagonal, all Gershgorin circles intersect the real line. Therefore, the circle theorem cannot establish the existence of complex eigenvalues.
answered Jan 30 at 8:01
PierreCarrePierreCarre
1,692212
$endgroup$
If the matrix has real entries, or at least real entries in the main diagonal, all Gershgorin circles intersect the real line. Therefore, the circle theorem cannot establish the existence of complex eigenvalues.
answered Jan 30 at 8:01
PierreCarrePierreCarre
1,692212
answered Jan 30 at 8:01
PierreCarrePierreCarre
1,692212
answered Jan 30 at 8:01
PierreCarrePierreCarre
1,692212
answered Jan 30 at 8:01
PierreCarrePierreCarre
1,692212
1,692212
$begingroup$
I think you are wrong, what about this example [0 -1 -1; 1 0 -1; 1 1 0] it has +/-i eigenvalues.
$endgroup$
– ChaosPredictor
Jan 30 at 17:41
$begingroup$
The matrix you mention has complex eigenvalues, sure, but what I say is that the theorem does not help to establish that fact. In the conditions I mentioned, the circles intersect the real line and therefore the theorem never rules out real eigenvalues.
$endgroup$
– PierreCarre
Jan 30 at 17:51
add a comment |
$begingroup$
I think you are wrong, what about this example [0 -1 -1; 1 0 -1; 1 1 0] it has +/-i eigenvalues.
$endgroup$
– ChaosPredictor
Jan 30 at 17:41
$begingroup$
The matrix you mention has complex eigenvalues, sure, but what I say is that the theorem does not help to establish that fact. In the conditions I mentioned, the circles intersect the real line and therefore the theorem never rules out real eigenvalues.
$endgroup$
– PierreCarre
Jan 30 at 17:51
$begingroup$
I think you are wrong, what about this example [0 -1 -1; 1 0 -1; 1 1 0] it has +/-i eigenvalues.
$endgroup$
– ChaosPredictor
Jan 30 at 17:41
$begingroup$
I think you are wrong, what about this example [0 -1 -1; 1 0 -1; 1 1 0] it has +/-i eigenvalues.
$endgroup$
– ChaosPredictor
Jan 30 at 17:41
$begingroup$
The matrix you mention has complex eigenvalues, sure, but what I say is that the theorem does not help to establish that fact. In the conditions I mentioned, the circles intersect the real line and therefore the theorem never rules out real eigenvalues.
$endgroup$
– PierreCarre
Jan 30 at 17:51
$begingroup$
The matrix you mention has complex eigenvalues, sure, but what I say is that the theorem does not help to establish that fact. In the conditions I mentioned, the circles intersect the real line and therefore the theorem never rules out real eigenvalues.
$endgroup$
– PierreCarre
Jan 30 at 17:51
add a comment |
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$begingroup$
If the matrix has complex entries, the eigenvalues can be complex, and not necessarily in conjugate pairs. For a complex matrix, if a Gershgorin circle or a union of more of them do not touch the x-axis, the existence of complex eigenvalue(s) is proven.
$endgroup$
– user376343
Jan 31 at 13:08