distinct eigenvalues
$begingroup$
When we say that we have ordered distinct eigenvalues. Does that mean that the the eigenvalues are decreasing $lambda_1 geq cdots geq lambda_p$ or strictly decreasing $lambda_1 > cdots > lambda_p$ ?
linear-algebra
asked Jun 7 '18 at 12:35
Mohammad MarrawiMohammad Marrawi
214
$endgroup$
add a comment |
$begingroup$
When we say that we have ordered distinct eigenvalues. Does that mean that the the eigenvalues are decreasing $lambda_1 geq cdots geq lambda_p$ or strictly decreasing $lambda_1 > cdots > lambda_p$ ?
linear-algebra
asked Jun 7 '18 at 12:35
Mohammad MarrawiMohammad Marrawi
214
$endgroup$
$begingroup$
We refer $geq$, not $>$.
$endgroup$
– Rafael Gonzalez Lopez
Jun 7 '18 at 12:37
add a comment |
$begingroup$
When we say that we have ordered distinct eigenvalues. Does that mean that the the eigenvalues are decreasing $lambda_1 geq cdots geq lambda_p$ or strictly decreasing $lambda_1 > cdots > lambda_p$ ?
linear-algebra
asked Jun 7 '18 at 12:35
Mohammad MarrawiMohammad Marrawi
214
$endgroup$
When we say that we have ordered distinct eigenvalues. Does that mean that the the eigenvalues are decreasing $lambda_1 geq cdots geq lambda_p$ or strictly decreasing $lambda_1 > cdots > lambda_p$ ?
linear-algebra
linear-algebra
asked Jun 7 '18 at 12:35
Mohammad MarrawiMohammad Marrawi
214
asked Jun 7 '18 at 12:35
Mohammad MarrawiMohammad Marrawi
214
asked Jun 7 '18 at 12:35
Mohammad MarrawiMohammad Marrawi
214
asked Jun 7 '18 at 12:35
Mohammad MarrawiMohammad Marrawi
214
asked Jun 7 '18 at 12:35
Mohammad MarrawiMohammad Marrawi
214
214
$begingroup$
We refer $geq$, not $>$.
$endgroup$
– Rafael Gonzalez Lopez
Jun 7 '18 at 12:37
add a comment |
$begingroup$
We refer $geq$, not $>$.
$endgroup$
– Rafael Gonzalez Lopez
Jun 7 '18 at 12:37
$begingroup$
We refer $geq$, not $>$.
$endgroup$
– Rafael Gonzalez Lopez
Jun 7 '18 at 12:37
$begingroup$
We refer $geq$, not $>$.
$endgroup$
– Rafael Gonzalez Lopez
Jun 7 '18 at 12:37
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
If the $lambda_i$ represent all of the eigenvalues, then the first notation is correct, since you will have equality where there are repeated eigenvalues and the order is not strict.
But if the $lambda_i$ represent only the distinct eigenvalues, then the second notation is correct, since distinct means that the repeats are not included and the order is strict.
For example, if the eigenvalues are $3, 3, 3, 2, 1, 1,$ then you would order the distinct eigenvalues as $3>2>1$. You could order all of the eigenvalues as $3geq 3geq 3geq 2geq 1geq 1$.
answered Jun 7 '18 at 12:45
MPWMPW
30.6k12157
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1 Answer
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$begingroup$
If the $lambda_i$ represent all of the eigenvalues, then the first notation is correct, since you will have equality where there are repeated eigenvalues and the order is not strict.
But if the $lambda_i$ represent only the distinct eigenvalues, then the second notation is correct, since distinct means that the repeats are not included and the order is strict.
For example, if the eigenvalues are $3, 3, 3, 2, 1, 1,$ then you would order the distinct eigenvalues as $3>2>1$. You could order all of the eigenvalues as $3geq 3geq 3geq 2geq 1geq 1$.
answered Jun 7 '18 at 12:45
MPWMPW
30.6k12157
$endgroup$
add a comment |
$begingroup$
If the $lambda_i$ represent all of the eigenvalues, then the first notation is correct, since you will have equality where there are repeated eigenvalues and the order is not strict.
But if the $lambda_i$ represent only the distinct eigenvalues, then the second notation is correct, since distinct means that the repeats are not included and the order is strict.
For example, if the eigenvalues are $3, 3, 3, 2, 1, 1,$ then you would order the distinct eigenvalues as $3>2>1$. You could order all of the eigenvalues as $3geq 3geq 3geq 2geq 1geq 1$.
answered Jun 7 '18 at 12:45
MPWMPW
30.6k12157
$endgroup$
add a comment |
$begingroup$
If the $lambda_i$ represent all of the eigenvalues, then the first notation is correct, since you will have equality where there are repeated eigenvalues and the order is not strict.
But if the $lambda_i$ represent only the distinct eigenvalues, then the second notation is correct, since distinct means that the repeats are not included and the order is strict.
For example, if the eigenvalues are $3, 3, 3, 2, 1, 1,$ then you would order the distinct eigenvalues as $3>2>1$. You could order all of the eigenvalues as $3geq 3geq 3geq 2geq 1geq 1$.
answered Jun 7 '18 at 12:45
MPWMPW
30.6k12157
$endgroup$
If the $lambda_i$ represent all of the eigenvalues, then the first notation is correct, since you will have equality where there are repeated eigenvalues and the order is not strict.
But if the $lambda_i$ represent only the distinct eigenvalues, then the second notation is correct, since distinct means that the repeats are not included and the order is strict.
For example, if the eigenvalues are $3, 3, 3, 2, 1, 1,$ then you would order the distinct eigenvalues as $3>2>1$. You could order all of the eigenvalues as $3geq 3geq 3geq 2geq 1geq 1$.
answered Jun 7 '18 at 12:45
MPWMPW
30.6k12157
answered Jun 7 '18 at 12:45
MPWMPW
30.6k12157
answered Jun 7 '18 at 12:45
MPWMPW
30.6k12157
answered Jun 7 '18 at 12:45
MPWMPW
30.6k12157
30.6k12157
add a comment |
add a comment |
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$begingroup$
We refer $geq$, not $>$.
$endgroup$
– Rafael Gonzalez Lopez
Jun 7 '18 at 12:37