This is the 16th day of my participation in the November Gwen Challenge. Check out the event details: The last Gwen Challenge 2021

This paper introduces the content related to quadratic derivative and extremum point under the defined problem.

Is there a point where the derivative is 0

  • For the general quadratic form:

f ( x ) = 1 2 x T A x – b T x + c \label 1 (1) f({\bf{x} }) = \frac{1}{2}{\bf{x^TAx} } – { {\bf{b} }^{\bf{T} } }{\bf{x} } + {\bf{c} } \tag{1} \label{1}
  • Derivative as follows:

f ‘ ( x ) = A x – b \label 2 (2) f'({\bf{x}}) = {\bf{Ax}} – {\bf{b}} \tag{2} \label{2}
  • The existence of a point whose derivative is 0 is equivalent to whether the equation Ax=b\bf{Ax}=\bf{b}Ax=b has a solution
  • Let rA=r(A){r_A} =r(A) rA=r(A) be the rank of A\bf{A}A
  • Ab\bf{Ab}Ab is the augmented matrix, rAb=r(Ab){r_{Ab}} =r(\bf{Ab})rAb=r(Ab) is the rank of the augmented matrix, where:
conditions conclusion

r A < r A b r_{A}<r_{Ab} 		The system of equations has no solution, and the quadratic form has no point whose derivative is 0

r A = r A b = n r_{A}=r_{Ab}=n 		The system has a unique solution, and the quadratic form has a unique point whose derivative is 0

r A = r A b < n r_{A}=r_{Ab}<n 		The system of equations has no array solution, the quadratic type has countless points whose derivative is 0

r A > r A b r_{A}>r_{Ab} 		No, the rank of the augmented matrix doesn’t get smaller

Whether an extreme point exists

  • When the point whose derivative is zero does not exist, that is, when the equations of \eqref2\eqref{2}\eqref2 have no solution, the extreme point does not exist

  • When the point whose derivative is 0 exists:

    • If A\bf{A}A is A positive definite matrix, then the formula \eqref1\eqref{1}\eqref1 has A minimum and is the minimum
    • If A\bf{A}A is A negative definite matrix, then the formula \eqref1\eqref{1}\eqref1 has A maximum and is the maximum
    • If A\bf{A}A is A semi-positive definite matrix, and there are eigenvalues of 0, since the premise is that the equations have solutions, then the rank of the augmented matrix is equal to the rank of the matrix A\bf{A}A, then the corresponding value in B \bf{b}b is 0, the equations have infinite sets of solutions, but some conditions need to be satisfied, in this set of conditions, the equations are minimal, In other words, if A\bf{A}A is A semi-positive definite matrix, \eqref1\eqref{1}\eqref1 has A minimum and is A minimum
    • In the same way, if A\bf{A}A is semi-negative definite matrix, then \eqref1\eqref{1}\eqref1 has the maximum value, that is, the maximum value
    • If the eigenvalues of A\bf{A}A are positive and negative, then \eqref1\eqref{1}\eqref1 has A minimum/maximum value, but not A minimum/maximum value, then \eqref1\eqref{1}\eqref1 has no minimum/maximum value

  • In order to make the discussion meaningful, the optimization of \eqref1\eqref{1}\eqref1 we will discuss later is to find its minimum value, that is, the minimum value, under the condition that A\bf{A}A is semi-positive definite matrix