How to check for local extrema or saddle point given an semidefinite matrix

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I've computed the Hessian of a given function $f(a,b,c) = y-asin(bx-c)$ and got the following result:

$beginpmatrix
0 & -xcdotcos(bx - c) & cos(bx - c) \
-xcdotcos(bx - c) & ax^2cdotsin(bx - c) & -axsin(bx - c) \
cos(bx - c) & -axcdotsin(bx - c) & acdotsin(bx - c)
endpmatrix$

This matrix is positive semi-definite and thus one can not state for a given point $P=(a_i,b_i,c_i)$ if it is a local min, max or a saddle point. Is there any other way to explicitly determine if we have a loc. min, max or saddle point?

analysis positive-semidefinite hessian-matrix

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edited Aug 2 at 15:28

Cornman

2,27521027

asked Aug 2 at 15:14

Daniyal

15317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Would you please tell us how $f$ is defined?
  – José Carlos Santos
  Aug 2 at 15:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh absolutely! I apologize for missing this. I'll edit my question.
  – Daniyal
  Aug 2 at 15:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  positive or negative semi definite ?
  – Ahmad Bazzi
  Aug 2 at 15:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  how have you determined the signature?
  – gimusi
  Aug 2 at 15:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AhmadBazzi, positive-semidefinite. I've set in this particular case x=4, y=3 and introduced as the point P = (1,3,2). Given x,y and P I computed the determinant for each of the leading principal minors
  – Daniyal
  Aug 2 at 15:24
  

  
  
  
  

 | 
show 2 more comments

up vote
2
down vote

favorite

I've computed the Hessian of a given function $f(a,b,c) = y-asin(bx-c)$ and got the following result:

$beginpmatrix
0 & -xcdotcos(bx - c) & cos(bx - c) \
-xcdotcos(bx - c) & ax^2cdotsin(bx - c) & -axsin(bx - c) \
cos(bx - c) & -axcdotsin(bx - c) & acdotsin(bx - c)
endpmatrix$

This matrix is positive semi-definite and thus one can not state for a given point $P=(a_i,b_i,c_i)$ if it is a local min, max or a saddle point. Is there any other way to explicitly determine if we have a loc. min, max or saddle point?

analysis positive-semidefinite hessian-matrix

share|cite|improve this question

edited Aug 2 at 15:28

Cornman

2,27521027

asked Aug 2 at 15:14

Daniyal

15317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Would you please tell us how $f$ is defined?
  – José Carlos Santos
  Aug 2 at 15:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh absolutely! I apologize for missing this. I'll edit my question.
  – Daniyal
  Aug 2 at 15:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  positive or negative semi definite ?
  – Ahmad Bazzi
  Aug 2 at 15:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  how have you determined the signature?
  – gimusi
  Aug 2 at 15:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AhmadBazzi, positive-semidefinite. I've set in this particular case x=4, y=3 and introduced as the point P = (1,3,2). Given x,y and P I computed the determinant for each of the leading principal minors
  – Daniyal
  Aug 2 at 15:24
  

  
  
  
  

 | 
show 2 more comments

up vote
2
down vote

favorite

up vote
2
down vote

favorite

I've computed the Hessian of a given function $f(a,b,c) = y-asin(bx-c)$ and got the following result:

$beginpmatrix
0 & -xcdotcos(bx - c) & cos(bx - c) \
-xcdotcos(bx - c) & ax^2cdotsin(bx - c) & -axsin(bx - c) \
cos(bx - c) & -axcdotsin(bx - c) & acdotsin(bx - c)
endpmatrix$

This matrix is positive semi-definite and thus one can not state for a given point $P=(a_i,b_i,c_i)$ if it is a local min, max or a saddle point. Is there any other way to explicitly determine if we have a loc. min, max or saddle point?

analysis positive-semidefinite hessian-matrix

share|cite|improve this question

edited Aug 2 at 15:28

Cornman

2,27521027

asked Aug 2 at 15:14

Daniyal

15317

I've computed the Hessian of a given function $f(a,b,c) = y-asin(bx-c)$ and got the following result:

$beginpmatrix
0 & -xcdotcos(bx - c) & cos(bx - c) \
-xcdotcos(bx - c) & ax^2cdotsin(bx - c) & -axsin(bx - c) \
cos(bx - c) & -axcdotsin(bx - c) & acdotsin(bx - c)
endpmatrix$

This matrix is positive semi-definite and thus one can not state for a given point $P=(a_i,b_i,c_i)$ if it is a local min, max or a saddle point. Is there any other way to explicitly determine if we have a loc. min, max or saddle point?

analysis positive-semidefinite hessian-matrix

share|cite|improve this question

edited Aug 2 at 15:28

Cornman

2,27521027

asked Aug 2 at 15:14

Daniyal

15317

share|cite|improve this question

share|cite|improve this question

edited Aug 2 at 15:28

Cornman

2,27521027

edited Aug 2 at 15:28

Cornman

2,27521027

edited Aug 2 at 15:28

Cornman

2,27521027

2,27521027

asked Aug 2 at 15:14

Daniyal

15317

asked Aug 2 at 15:14

Daniyal

15317

asked Aug 2 at 15:14

Daniyal

15317

15317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Would you please tell us how $f$ is defined?
  – José Carlos Santos
  Aug 2 at 15:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh absolutely! I apologize for missing this. I'll edit my question.
  – Daniyal
  Aug 2 at 15:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  positive or negative semi definite ?
  – Ahmad Bazzi
  Aug 2 at 15:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  how have you determined the signature?
  – gimusi
  Aug 2 at 15:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AhmadBazzi, positive-semidefinite. I've set in this particular case x=4, y=3 and introduced as the point P = (1,3,2). Given x,y and P I computed the determinant for each of the leading principal minors
  – Daniyal
  Aug 2 at 15:24
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Would you please tell us how $f$ is defined?
  – José Carlos Santos
  Aug 2 at 15:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh absolutely! I apologize for missing this. I'll edit my question.
  – Daniyal
  Aug 2 at 15:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  positive or negative semi definite ?
  – Ahmad Bazzi
  Aug 2 at 15:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  how have you determined the signature?
  – gimusi
  Aug 2 at 15:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AhmadBazzi, positive-semidefinite. I've set in this particular case x=4, y=3 and introduced as the point P = (1,3,2). Given x,y and P I computed the determinant for each of the leading principal minors
  – Daniyal
  Aug 2 at 15:24
  

  
  
  
  

Would you please tell us how $f$ is defined?
– José Carlos Santos
Aug 2 at 15:15

Would you please tell us how $f$ is defined?
– José Carlos Santos
Aug 2 at 15:15

oh absolutely! I apologize for missing this. I'll edit my question.
– Daniyal
Aug 2 at 15:16

oh absolutely! I apologize for missing this. I'll edit my question.
– Daniyal
Aug 2 at 15:16

positive or negative semi definite ?
– Ahmad Bazzi
Aug 2 at 15:18

positive or negative semi definite ?
– Ahmad Bazzi
Aug 2 at 15:18

how have you determined the signature?
– gimusi
Aug 2 at 15:20

how have you determined the signature?
– gimusi
Aug 2 at 15:20

@AhmadBazzi, positive-semidefinite. I've set in this particular case x=4, y=3 and introduced as the point P = (1,3,2). Given x,y and P I computed the determinant for each of the leading principal minors
– Daniyal
Aug 2 at 15:24

@AhmadBazzi, positive-semidefinite. I've set in this particular case x=4, y=3 and introduced as the point P = (1,3,2). Given x,y and P I computed the determinant for each of the leading principal minors
– Daniyal
Aug 2 at 15:24

 | 
show 2 more comments

1 Answer
1

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0
down vote

With reference to the given matrix, we have that

$$det(asin(bx - c))=asin(bx - c)$$

$$beginvmatrix
0 & -xcos(bx - c) \
-xcos(bx - c) & ax^2sin(bx - c) \
endvmatrix=-x^2cos^2(bx - c)$$

$$beginvmatrix
0 & -xcdotcos(bx - c) & cos(bx - c) \
-xcdotcos(bx - c) & ax^2cdotsin(bx - c) & -axsin(bx - c) \
cos(bx - c) & -axcdotsin(bx - c) & acdotsin(bx - c)
endvmatrix=$$
$$=xcdotcos(bx - c)(-axcos(bx - c)sin(bx - c)+axcos(bx - c)sin(bx - c))+cos(bx - c)(ax^2cos(bx - c)sin(bx - c)-ax^2cos(bx - c)sin(bx - c))=0$$

therefore for $x^2cos^2(bx - c)neq 0$

• $asin(bx - c)le 0$ the matrix is negative semidefinite




• $asin(bx - c)>0$ the matrix is indefinite

and for $x^2cos^2(bx - c)= 0$

• $asin(bx - c)< 0$ the matrix is negative semidefinite




• $asin(bx - c)>0$ the matrix is positive semidefinite

Note that the condition for critical points implies

• $f_a=-sin(bx - c)=0implies sin(bx - c)=0$




• $f_b=-axsin(bx - c)=0implies sin(bx - c)=0 lor x=0$




• $f_c=asin(bx - c)=0implies sin(bx - c)=0$

therefore

• $cos(bx - c)=pm 1$

and the expression for the Hessian simplifies.

share|cite|improve this answer

edited Aug 2 at 17:46

answered Aug 2 at 15:43

gimusi

63.8k73480

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

With reference to the given matrix, we have that

$$det(asin(bx - c))=asin(bx - c)$$

$$beginvmatrix
0 & -xcos(bx - c) \
-xcos(bx - c) & ax^2sin(bx - c) \
endvmatrix=-x^2cos^2(bx - c)$$

$$beginvmatrix
0 & -xcdotcos(bx - c) & cos(bx - c) \
-xcdotcos(bx - c) & ax^2cdotsin(bx - c) & -axsin(bx - c) \
cos(bx - c) & -axcdotsin(bx - c) & acdotsin(bx - c)
endvmatrix=$$
$$=xcdotcos(bx - c)(-axcos(bx - c)sin(bx - c)+axcos(bx - c)sin(bx - c))+cos(bx - c)(ax^2cos(bx - c)sin(bx - c)-ax^2cos(bx - c)sin(bx - c))=0$$

therefore for $x^2cos^2(bx - c)neq 0$

• $asin(bx - c)le 0$ the matrix is negative semidefinite




• $asin(bx - c)>0$ the matrix is indefinite

and for $x^2cos^2(bx - c)= 0$

• $asin(bx - c)< 0$ the matrix is negative semidefinite




• $asin(bx - c)>0$ the matrix is positive semidefinite

Note that the condition for critical points implies

• $f_a=-sin(bx - c)=0implies sin(bx - c)=0$




• $f_b=-axsin(bx - c)=0implies sin(bx - c)=0 lor x=0$




• $f_c=asin(bx - c)=0implies sin(bx - c)=0$

therefore

• $cos(bx - c)=pm 1$

and the expression for the Hessian simplifies.

share|cite|improve this answer

edited Aug 2 at 17:46

answered Aug 2 at 15:43

gimusi

63.8k73480

add a comment | 

up vote
0
down vote

With reference to the given matrix, we have that

$$det(asin(bx - c))=asin(bx - c)$$

$$beginvmatrix
0 & -xcos(bx - c) \
-xcos(bx - c) & ax^2sin(bx - c) \
endvmatrix=-x^2cos^2(bx - c)$$

$$beginvmatrix
0 & -xcdotcos(bx - c) & cos(bx - c) \
-xcdotcos(bx - c) & ax^2cdotsin(bx - c) & -axsin(bx - c) \
cos(bx - c) & -axcdotsin(bx - c) & acdotsin(bx - c)
endvmatrix=$$
$$=xcdotcos(bx - c)(-axcos(bx - c)sin(bx - c)+axcos(bx - c)sin(bx - c))+cos(bx - c)(ax^2cos(bx - c)sin(bx - c)-ax^2cos(bx - c)sin(bx - c))=0$$

therefore for $x^2cos^2(bx - c)neq 0$

• $asin(bx - c)le 0$ the matrix is negative semidefinite




• $asin(bx - c)>0$ the matrix is indefinite

and for $x^2cos^2(bx - c)= 0$

• $asin(bx - c)< 0$ the matrix is negative semidefinite




• $asin(bx - c)>0$ the matrix is positive semidefinite

Note that the condition for critical points implies

• $f_a=-sin(bx - c)=0implies sin(bx - c)=0$




• $f_b=-axsin(bx - c)=0implies sin(bx - c)=0 lor x=0$




• $f_c=asin(bx - c)=0implies sin(bx - c)=0$

therefore

• $cos(bx - c)=pm 1$

and the expression for the Hessian simplifies.

share|cite|improve this answer

edited Aug 2 at 17:46

answered Aug 2 at 15:43

gimusi

63.8k73480

add a comment | 

up vote
0
down vote

up vote
0
down vote

With reference to the given matrix, we have that

$$det(asin(bx - c))=asin(bx - c)$$

$$beginvmatrix
0 & -xcos(bx - c) \
-xcos(bx - c) & ax^2sin(bx - c) \
endvmatrix=-x^2cos^2(bx - c)$$

$$beginvmatrix
0 & -xcdotcos(bx - c) & cos(bx - c) \
-xcdotcos(bx - c) & ax^2cdotsin(bx - c) & -axsin(bx - c) \
cos(bx - c) & -axcdotsin(bx - c) & acdotsin(bx - c)
endvmatrix=$$
$$=xcdotcos(bx - c)(-axcos(bx - c)sin(bx - c)+axcos(bx - c)sin(bx - c))+cos(bx - c)(ax^2cos(bx - c)sin(bx - c)-ax^2cos(bx - c)sin(bx - c))=0$$

therefore for $x^2cos^2(bx - c)neq 0$

• $asin(bx - c)le 0$ the matrix is negative semidefinite




• $asin(bx - c)>0$ the matrix is indefinite

and for $x^2cos^2(bx - c)= 0$

• $asin(bx - c)< 0$ the matrix is negative semidefinite




• $asin(bx - c)>0$ the matrix is positive semidefinite

Note that the condition for critical points implies

• $f_a=-sin(bx - c)=0implies sin(bx - c)=0$




• $f_b=-axsin(bx - c)=0implies sin(bx - c)=0 lor x=0$




• $f_c=asin(bx - c)=0implies sin(bx - c)=0$

therefore

• $cos(bx - c)=pm 1$

and the expression for the Hessian simplifies.

share|cite|improve this answer

edited Aug 2 at 17:46

answered Aug 2 at 15:43

gimusi

63.8k73480

With reference to the given matrix, we have that

$$det(asin(bx - c))=asin(bx - c)$$

$$beginvmatrix
0 & -xcos(bx - c) \
-xcos(bx - c) & ax^2sin(bx - c) \
endvmatrix=-x^2cos^2(bx - c)$$

$$beginvmatrix
0 & -xcdotcos(bx - c) & cos(bx - c) \
-xcdotcos(bx - c) & ax^2cdotsin(bx - c) & -axsin(bx - c) \
cos(bx - c) & -axcdotsin(bx - c) & acdotsin(bx - c)
endvmatrix=$$
$$=xcdotcos(bx - c)(-axcos(bx - c)sin(bx - c)+axcos(bx - c)sin(bx - c))+cos(bx - c)(ax^2cos(bx - c)sin(bx - c)-ax^2cos(bx - c)sin(bx - c))=0$$

therefore for $x^2cos^2(bx - c)neq 0$

• $asin(bx - c)le 0$ the matrix is negative semidefinite




• $asin(bx - c)>0$ the matrix is indefinite

and for $x^2cos^2(bx - c)= 0$

• $asin(bx - c)< 0$ the matrix is negative semidefinite




• $asin(bx - c)>0$ the matrix is positive semidefinite

Note that the condition for critical points implies

• $f_a=-sin(bx - c)=0implies sin(bx - c)=0$




• $f_b=-axsin(bx - c)=0implies sin(bx - c)=0 lor x=0$




• $f_c=asin(bx - c)=0implies sin(bx - c)=0$

therefore

• $cos(bx - c)=pm 1$

and the expression for the Hessian simplifies.

share|cite|improve this answer

edited Aug 2 at 17:46

answered Aug 2 at 15:43

gimusi

63.8k73480

share|cite|improve this answer

share|cite|improve this answer

edited Aug 2 at 17:46

edited Aug 2 at 17:46

edited Aug 2 at 17:46

answered Aug 2 at 15:43

gimusi

63.8k73480

answered Aug 2 at 15:43

gimusi

63.8k73480

answered Aug 2 at 15:43

gimusi

63.8k73480

63.8k73480

add a comment | 

add a comment | 

 

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