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Why is positive definiteness required for a global minimum to exist?

0

$begingroup$

When we optimize

$$min x'Ax$$

why does matrix $A$ need to be positive definite in order to have a global minimum?

matrices optimization convex-optimization positive-definite quadratic-programming

share|cite|improve this question

edited Mar 23 at 9:22

Rodrigo de Azevedo

13.2k41962

asked Mar 23 at 3:13

JOHNJOHN

183

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is not required. If $A=0$ then there is a global minimum. You need positive semi definite. Otherwise there is a direction in which you can produce values that are unbounded below.
  $endgroup$
  – copper.hat
  Mar 23 at 3:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @copper.hat what is the requirement to have a unique solution?
  $endgroup$
  – JOHN
  Mar 23 at 3:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you guess? Where is the derivative (of a convex function) zero?
  $endgroup$
  – copper.hat
  Mar 23 at 3:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @copper.hat you would want $Ax=0$ has a solution, so the det𝐀 != 0. and second derivative to be negative. I don't know how could A be expressed to meet this.
  $endgroup$
  – JOHN
  Mar 23 at 3:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the second derivative of $x mapsto x'Ax$??
  $endgroup$
  – copper.hat
  Mar 23 at 3:29
  

  
  
  
  

 | 
show 4 more comments

0

$begingroup$

When we optimize

$$min x'Ax$$

why does matrix $A$ need to be positive definite in order to have a global minimum?

matrices optimization convex-optimization positive-definite quadratic-programming

share|cite|improve this question

edited Mar 23 at 9:22

Rodrigo de Azevedo

13.2k41962

asked Mar 23 at 3:13

JOHNJOHN

183

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is not required. If $A=0$ then there is a global minimum. You need positive semi definite. Otherwise there is a direction in which you can produce values that are unbounded below.
  $endgroup$
  – copper.hat
  Mar 23 at 3:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @copper.hat what is the requirement to have a unique solution?
  $endgroup$
  – JOHN
  Mar 23 at 3:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you guess? Where is the derivative (of a convex function) zero?
  $endgroup$
  – copper.hat
  Mar 23 at 3:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @copper.hat you would want $Ax=0$ has a solution, so the det𝐀 != 0. and second derivative to be negative. I don't know how could A be expressed to meet this.
  $endgroup$
  – JOHN
  Mar 23 at 3:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the second derivative of $x mapsto x'Ax$??
  $endgroup$
  – copper.hat
  Mar 23 at 3:29
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

When we optimize

$$min x'Ax$$

why does matrix $A$ need to be positive definite in order to have a global minimum?

matrices optimization convex-optimization positive-definite quadratic-programming

share|cite|improve this question

edited Mar 23 at 9:22

Rodrigo de Azevedo

13.2k41962

asked Mar 23 at 3:13

JOHNJOHN

183

$endgroup$

share|cite|improve this question

edited Mar 23 at 9:22

Rodrigo de Azevedo

13.2k41962

asked Mar 23 at 3:13

JOHNJOHN

183

share|cite|improve this question

edited Mar 23 at 9:22

Rodrigo de Azevedo

13.2k41962

asked Mar 23 at 3:13

JOHNJOHN

183

edited Mar 23 at 9:22

Rodrigo de Azevedo

13.2k41962

edited Mar 23 at 9:22

Rodrigo de Azevedo

13.2k41962

asked Mar 23 at 3:13

JOHNJOHN

183

asked Mar 23 at 3:13

JOHNJOHN

183