The condition of the positiveness of a ordinary quadratic form can be derived by getting the condition of positiveness of a square matrix, like ${v}^{T}{A}{v} \geq 0$ is equal to matrix $A \geq 0$ where $v \in \mathbb{R}^3$.
And what is the condition of positiveness if vector $v$ has a specified form?
If $v = [x^2\quad y^2\quad z^2]$ and $x,y,z \in \mathbb{R}$, my question is what condition matrix $A$ should satisfy so that a quadratic form, $v^{T}Av$, is positive $\forall x,y,z \in \mathbb{R}$?
And I think in this case, matrix $A$ does not have to be positive to make this specified quadratic form positive. Positiveness of matrix $A$ is a necessary but not sufficient condition. I was wondering if there is a necessary and sufficient condition for this problem.
Thanks.
My guess is that we can find only sufficient conditions for matrix $A$.
If we set $A = \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33}\end{bmatrix} $, then we have:
$v^T A v= x^4A_{11} + y^4A_{22} + z^4A_{33} + x^2y^2(A_{21} + A_{12}) + x^2z^2(A_{13} + A_{31}) + y^2z^2(A_{23} + A_{32})\tag{1},$
where $v^T = [ x^2 \quad y^2 \quad z^2]$.
Some sufficient conditions for matrix $A$, so that $v^T A v \ge 0$:
From $(1)$ we can easily figure out that if $A$ is a skew-symmetric-like matrix (which means $A_{ij} = -A_{ij}, \forall i\neq j$), then $$v^T A v = x^4A_{11} + y^4A_{22} + z^4A_{33}.$$ So, we can conclude that if all the diagonal elements are non-negative, then $v^TAv \ge 0,\forall x,y,z \in \mathbb R$.