Understanding negative definite/semidefinite functions [closed]
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I'm working on control theory and have some difficulty understanding if a function is negative definite or semidefinite.
Given the system
$dot{x_1} = -x_2^2$
$dot{x_2} = -x_1^2x_2 + x_1^3 - x_2$
How do I determine if
$-x_1^2x_2^4 - x_2^4$
is negative definite or negative semidefinite?
linear-algebra ordinary-differential-equations control-theory
edited Dec 13 '18 at 16:24
Omnomnomnom
129k794188
asked Dec 13 '18 at 16:16
fendrbudfendrbud
11
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closed as off-topic by Brahadeesh, Federico, jameselmore, Davide Giraudo, Namaste Dec 13 '18 at 22:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, Federico, jameselmore, Davide Giraudo, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I'm working on control theory and have some difficulty understanding if a function is negative definite or semidefinite.
Given the system
$dot{x_1} = -x_2^2$
$dot{x_2} = -x_1^2x_2 + x_1^3 - x_2$
How do I determine if
$-x_1^2x_2^4 - x_2^4$
is negative definite or negative semidefinite?
linear-algebra ordinary-differential-equations control-theory
edited Dec 13 '18 at 16:24
Omnomnomnom
129k794188
asked Dec 13 '18 at 16:16
fendrbudfendrbud
11
$endgroup$
closed as off-topic by Brahadeesh, Federico, jameselmore, Davide Giraudo, Namaste Dec 13 '18 at 22:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, Federico, jameselmore, Davide Giraudo, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
"negative-definite" as applied to a function has different meaning in different contexts. Is this definition the one you're working with?
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– Omnomnomnom
Dec 13 '18 at 16:27
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Take $x_1 = 1$ and $x_2 = 0$. Then clearly your expression equals $0$ although $(x_1, x_2) = (1, 0) neq (0, 0)$. Therefore, your expression is negative semi-definite.
$endgroup$
– SampleTime
Dec 13 '18 at 21:31
add a comment |
$begingroup$
I'm working on control theory and have some difficulty understanding if a function is negative definite or semidefinite.
Given the system
$dot{x_1} = -x_2^2$
$dot{x_2} = -x_1^2x_2 + x_1^3 - x_2$
How do I determine if
$-x_1^2x_2^4 - x_2^4$
is negative definite or negative semidefinite?
linear-algebra ordinary-differential-equations control-theory
edited Dec 13 '18 at 16:24
Omnomnomnom
129k794188
asked Dec 13 '18 at 16:16
fendrbudfendrbud
11
$endgroup$
I'm working on control theory and have some difficulty understanding if a function is negative definite or semidefinite.
Given the system
$dot{x_1} = -x_2^2$
$dot{x_2} = -x_1^2x_2 + x_1^3 - x_2$
How do I determine if
$-x_1^2x_2^4 - x_2^4$
is negative definite or negative semidefinite?
linear-algebra ordinary-differential-equations control-theory
linear-algebra ordinary-differential-equations control-theory
edited Dec 13 '18 at 16:24
Omnomnomnom
129k794188
asked Dec 13 '18 at 16:16
fendrbudfendrbud
11
edited Dec 13 '18 at 16:24
Omnomnomnom
129k794188
asked Dec 13 '18 at 16:16
fendrbudfendrbud
11
edited Dec 13 '18 at 16:24
Omnomnomnom
129k794188
edited Dec 13 '18 at 16:24
Omnomnomnom
129k794188
edited Dec 13 '18 at 16:24
Omnomnomnom
129k794188
129k794188
asked Dec 13 '18 at 16:16
fendrbudfendrbud
11
asked Dec 13 '18 at 16:16
fendrbudfendrbud
11
asked Dec 13 '18 at 16:16
fendrbudfendrbud
11
11
closed as off-topic by Brahadeesh, Federico, jameselmore, Davide Giraudo, Namaste Dec 13 '18 at 22:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, Federico, jameselmore, Davide Giraudo, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Brahadeesh, Federico, jameselmore, Davide Giraudo, Namaste Dec 13 '18 at 22:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, Federico, jameselmore, Davide Giraudo, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
"negative-definite" as applied to a function has different meaning in different contexts. Is this definition the one you're working with?
$endgroup$
– Omnomnomnom
Dec 13 '18 at 16:27
$begingroup$
Take $x_1 = 1$ and $x_2 = 0$. Then clearly your expression equals $0$ although $(x_1, x_2) = (1, 0) neq (0, 0)$. Therefore, your expression is negative semi-definite.
$endgroup$
– SampleTime
Dec 13 '18 at 21:31
add a comment |
1
$begingroup$
"negative-definite" as applied to a function has different meaning in different contexts. Is this definition the one you're working with?
$endgroup$
– Omnomnomnom
Dec 13 '18 at 16:27
$begingroup$
Take $x_1 = 1$ and $x_2 = 0$. Then clearly your expression equals $0$ although $(x_1, x_2) = (1, 0) neq (0, 0)$. Therefore, your expression is negative semi-definite.
$endgroup$
– SampleTime
Dec 13 '18 at 21:31
1
1
$begingroup$
"negative-definite" as applied to a function has different meaning in different contexts. Is this definition the one you're working with?
$endgroup$
– Omnomnomnom
Dec 13 '18 at 16:27
$begingroup$
"negative-definite" as applied to a function has different meaning in different contexts. Is this definition the one you're working with?
$endgroup$
– Omnomnomnom
Dec 13 '18 at 16:27
$begingroup$
Take $x_1 = 1$ and $x_2 = 0$. Then clearly your expression equals $0$ although $(x_1, x_2) = (1, 0) neq (0, 0)$. Therefore, your expression is negative semi-definite.
$endgroup$
– SampleTime
Dec 13 '18 at 21:31
$begingroup$
Take $x_1 = 1$ and $x_2 = 0$. Then clearly your expression equals $0$ although $(x_1, x_2) = (1, 0) neq (0, 0)$. Therefore, your expression is negative semi-definite.
$endgroup$
– SampleTime
Dec 13 '18 at 21:31
add a comment |
1 Answer
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Is there $(x_1, x_2) neq (0,0)$ such that the function attains $0$? If so, it is semidefinite.
Note that I don't know your definition, I'm just guessing by the usual definition for bilinear forms.
answered Dec 13 '18 at 16:26
StockfishStockfish
62726
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Is there $(x_1, x_2) neq (0,0)$ such that the function attains $0$? If so, it is semidefinite.
Note that I don't know your definition, I'm just guessing by the usual definition for bilinear forms.
answered Dec 13 '18 at 16:26
StockfishStockfish
62726
$endgroup$
add a comment |
$begingroup$
Is there $(x_1, x_2) neq (0,0)$ such that the function attains $0$? If so, it is semidefinite.
Note that I don't know your definition, I'm just guessing by the usual definition for bilinear forms.
answered Dec 13 '18 at 16:26
StockfishStockfish
62726
$endgroup$
add a comment |
$begingroup$
Is there $(x_1, x_2) neq (0,0)$ such that the function attains $0$? If so, it is semidefinite.
Note that I don't know your definition, I'm just guessing by the usual definition for bilinear forms.
answered Dec 13 '18 at 16:26
StockfishStockfish
62726
$endgroup$
Is there $(x_1, x_2) neq (0,0)$ such that the function attains $0$? If so, it is semidefinite.
Note that I don't know your definition, I'm just guessing by the usual definition for bilinear forms.
answered Dec 13 '18 at 16:26
StockfishStockfish
62726
answered Dec 13 '18 at 16:26
StockfishStockfish
62726
answered Dec 13 '18 at 16:26
StockfishStockfish
62726
answered Dec 13 '18 at 16:26
StockfishStockfish
62726
62726
add a comment |
add a comment |
1
$begingroup$
"negative-definite" as applied to a function has different meaning in different contexts. Is this definition the one you're working with?
$endgroup$
– Omnomnomnom
Dec 13 '18 at 16:27
$begingroup$
Take $x_1 = 1$ and $x_2 = 0$. Then clearly your expression equals $0$ although $(x_1, x_2) = (1, 0) neq (0, 0)$. Therefore, your expression is negative semi-definite.
$endgroup$
– SampleTime
Dec 13 '18 at 21:31