# Determining positive/negative definite of quadratic form using Hessian matrix method?

The matrix A is given as:

$$\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}$$

Given that the domain of the quadratic form $$x^T Ax$$ is restricted as the following,

$$D={x\in R^3, x_1+x_2+x_3=0}$$

determine whether the quadratic form is positive/negative definite or positive/negative semidefinite.

I know how to solve this kind of problem when there's no restriction/constraint, but I have no idea how to determine this when there is a restriction. The textbook talks about some bordered Hessian matrix method and some other Hessian matrix method but I have no idea how to apply these methods...

$$A=\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix} = \mathbf {11}^T-I$$

and you want to know the signature with the added restriction that we only consider $$\mathbf x$$ such that $$\mathbf 1^T\mathbf x = 0$$.

There are a few different approaches here. Probably the simplest is to make use of the projection matrix
$$P:= I-\alpha\mathbf {11}^T$$, where $$\alpha$$ is selected so $$P$$ has rank $$n-1$$, i.e. $$\alpha =\Big(\mathbf 1^T \mathbf 1\Big)^{-1} = n^{-1}$$
check: $$P^2=P=P^T$$

$$B:= P^TAP$$ and notice that if $$\mathbf x \perp \mathbf 1$$
$$P\mathbf x = \mathbf x \implies\mathbf x^T B \mathbf x = \mathbf x^T A \mathbf x$$

$$B$$ is real symmetric with $$\mathbf 1$$ in its kernel so you may in general orthogonally diagonalize it and consider the $$n-1$$ orthogonal eigenvvectors of interest to figure out its signature (or run $$LDL^T$$ factorization, etc.)

That said, your problem is particularly simple:
$$B= P^TAP = P^T\big(\mathbf {11}^T-I\big)P = -P^TP = -P$$
since $$-P$$ has signature $$\big(0,n-1\big)$$ with only vectors $$\propto\mathbf 1$$ in its kernel this tells you that $$\mathbf x^T A\mathbf x \lt 0$$ for any non-zero $$\mathbf x \perp \mathbf 1$$

• Have you an idea about the "Hessian matrix method" the asker is referring to?
– user
Sep 16, 2020 at 21:16
• @user the Bordered Hessian method would be a little tedious but feasible here. Idea is set up a Lagrangian in the normal way -- try to solve for a maximum (implies negative definite). Compute gradient, set = 0, solve, then compute Hessian as normal. Then stick that Hessian as the main block of the bigger bordered Hessian. There's some finessing so the constraint doesn't vanish under the 2nd derivative. pages 5-7 of this are relevant mro.massey.ac.nz/handle/10179/4457 . OPs matrix looks like something from MDS Schoenberg theory and using projectors is much simpler. Sep 16, 2020 at 23:49
• Thanks! But whst about using the inequality as I’ve proposed? Is there something wrong with thst?
– user
Sep 17, 2020 at 0:06
• @user it seems fine to me. I can only guess at some of the down votes on this site. Yet another approach to this problem would be to place it in a Cayley Menger matrix (discussed in Dattoro's book : ccrma.stanford.edu/~dattorro/mybook.html ) Sep 17, 2020 at 0:12
• Thanks for the information and material. I was wondering for some misunderstanding in the question or other issues.
– user
Sep 17, 2020 at 0:17

We have that

$$x^T Ax =2x_1x_2+2x_2x_3+2x_3x_1$$

and for any $$x=(x_1,x_2,x_3)\neq 0$$

$$(x_1+x_2+x_3)^2=0 \implies 2x_1x_2+2x_2x_3+2x_3x_1 =-(x_1^2+x_2^2+x_3^2)<0$$

therefore by this restriction the quadratic form is negative definite.