If a matrix $G$ is positive semidefinite on hyperplane, then it is positive semidefinite on whole space

Question

I'm trying to show this statement:

Let $G = 1w^T + w1^T - \frac{1}{2}D$ , where $w$ is nonnegative, $z^TDz \leq 0$ for all $z$ such that $1^Tz=0$ and $1^TD1 \geq 0$.

Show that if $z^TGz\geq 0$ for all $z$ such that $1^Tz = 0$, then $z^TGz \geq 0$ for all $z \in \Bbb R^n$.

Thanks for any advice.

Answer 1

Hint:

Let $H = \{v \in \mathbb{R}^{n-1} \mid 1^Tv = 0\}$. $H$ is an $n-1$ dimensional subspace of $\mathbb{R}^{n-1}$ (a basis for it is $(1,-1,\ldots), (0,1,-1,\ldots)$, etc.)

The vectors orthogonal to $H$ are precisely the scalar multiples of $1$.
Thus any $z \in \mathbb{R}^{n}$ decomposes as $z = v + a1$ with $a \in \mathbb{R}$ and $1^Tv = 0$.

If you play with that decomposition and your definition of $G$, the result falls out quickly.