Question in Proof of second order condition with linear constraints

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This discussion focuses on the proof of sign definiteness of quadratic forms with linear constraints, specifically utilizing the bordered Hessian method. The quadratic form "E" is derived from the Hessian of the objective function and is structured to integrate with another quadratic form presented later in the proof. The conditions for positive and negative definiteness are established through the determinant relationships involving the bordered Hessian and the quadratic form "E," particularly the expression (-1)^(j−k) det(B1)^2 det(E). The manipulation of the basis of the bordered Hessian is crucial for determining the sign definiteness of the quadratic form "Q."

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holemole
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http://www.math.northwestern.edu/~clark/285/2006-07/handouts/lin-constraint.pdf


It's actually proof of finding sign definiteness of quadratic form with linear constraints with sign of submatrices of bordered hessian.

The proof is from page 2~page 3. I have 2 questions:

1. From about 6th line of the proof it mentions "E" being a quadratic form of A, hessian of our objective function. Its specific form is mentioned in the paper, but why is it formed that way? Is it just to make a quadratic form that will fit in another quadratic form presented later in the proof? I have similar question with quadratic form of H, the bordered hessian.

2. The last 6lines of the proof.

The two conditions each representing positive and negative definite case, as far as I understand, follows from (-1)^k det(B1)^2 det(E). so in the negative definite case where does (-1)^(j−k) et(Hj) = (-1)^(j−2k) det(B1)^2 det(Ej−2k) > 0 come from?
 

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Kind of figured it out by myself now. Reason why the proof states E is because while E is quadratic form of A, Q is again quadratic form of E. Thus sign definiteness of Q can rely on det(E), which is attainable if we follow the proof's manipulation ofchanging the basis of the bordered hessian.

the (-1)^(j-2k) comes from the fact that in order for Q to be negative semi-definite its discriminant(in this case det(E)) needs to have negative-positive-negative...signs for its submatrices. (-1)^(j-2k) allows this, thus multiplied to both sides of the equation.
 

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