Sylvester's Matrix Theorem for n=3

[pvortex]

I guess this is more of a linear algebra question, but I need to do it in the frame of calculus. (So I guess that makes it a really hard algebra question).

Basically, for a three-dimensional Hessian of the form Hf=[MATH]\frac{1}{2}[/MATH]x₀^TBx₀, I need to prove that the function (or more accurately its associated quadratic function) is positive definite if the 3x3 diagonal matrix B is positive definite. That is, the determinants of the submatrices of B are all positive.

I have matrix B as [MATH] \begin{bmatrix} a & b & d \\ b & c & e \\ d & e & f \end{bmatrix} [/MATH].

For conditions, I have [MATH]a > 0[/MATH], [MATH]ac-b^2 > 0[/MATH], and [MATH]acf-ae^2-b^2f-cd^2 > 0[/MATH].

From completing the square a few times, I've managed to get [MATH]\frac{1}{2}[a(x+\frac{b}{a}y)^2+(c-\frac{b^2}{a})(y+\frac{e}{c-\frac{b^2}{a}}z)^2+A(z+\frac{d}{A}x)^2-\frac{d^2}{A}x^2][/MATH]where [MATH]A=f-\frac{e^2}{c-\frac{b^2}{a}}=\frac{acf-ae^2-b^2f}{ac-b^2}[/MATH].

I've proven the first two conditions (submatrix determinants) and part of the third and final condition, but I can't seem to get [MATH]cd^2[/MATH] in a working inequality no matter what I do. Any and all help is obliged.

 

[pvortex]

I figure I might as well answer my own question.

From the polynomial and using the property of a square to always be positive, we know that [MATH]a>0[/MATH] from [MATH]a(x+\frac{b}{a}y)^2 [/MATH] in order for the polynomial to be positive definite.

We know [MATH]ac-b^2>0[/MATH] from the fact that [MATH]c-\frac{b^2}{a}[/MATH] from the coefficient of the second squared term must be positive ([MATH]c-\frac{b^2}{a}>0\Rightarrow ac-b^2>0[/MATH])

I am unsure of the validity of this part, but we know that [MATH]\frac{acf-ae^2-b^2f}{ac-b^2}>0[/MATH] from the last squared term. As [MATH]ac-b^2>0[/MATH], therefore [MATH]acf-ae^2-b^2f>0[/MATH]. Thus [MATH]acf-ae^2-b^2f-cd^2>-cd^2[/MATH]. Also from [MATH]ac-b^2>0[/MATH] and knowing [MATH]a>0[/MATH], as [MATH]b^2>0[/MATH], [MATH]c>0[/MATH]. And as [MATH]d^2>0[/MATH], therefore [MATH]-cd^2<0[/MATH]. Thus to be positive definite, [MATH]acf-ae^2-b^2f-cd^2>0[/MATH].