I need your help with proof of this equivalence.
Let $V = I - \frac{1}{n}\bf{11^T}$ and $z^TDz <= 0 \forall z: \bf{1}^Tz = 0$. Show that following two statements are equivalent:
(i) $-VDV \succeq 0$,
(ii) $z^TVDVz <= 0, \forall z: \bf{1}^Tz = 0$.
I have shown that (i) implies (ii), but I have problem with the other direction.
I'd be grateful for any advice.
We must show that $(i)\to(ii)$ and $(ii)\to(i)$. $(i)\to(ii)$ is true according to definition. To show $(ii)\to (i)$ we assume $(ii)$ and proceed through. Since $(ii)$ is assumed to be true we have $$z^TVDVz <= 0, \forall z: \bf{1}^Tz = 0$$ note that the following set of vectors span whole the $\Bbb R^n$:$$\forall1\le i\le n-1\qquad,\qquad r_i=(10\cdots0\ \ \underbrace{-1}_{\text{(i+1)th place}} \ \ 0\cdots0)^T\\r_n=(10\cdots0)^T$$(why?) then positive definition of $VDV$ is equivalent to the following$$\forall 1\le i\le n\qquad,\qquad r_i^TVDVr_i\le 0$$for $1\le i\le n-1$ those equations hold according to $(ii)$. It remains to show that $$r_n^TVDVr_n\le 0$$we know that $$W=Vr_n=\begin{bmatrix}1-\dfrac{1}{n}\\-\dfrac{1}{n}\\.\\.\\.\\-\dfrac{1}{n}\end{bmatrix}$$therefore $$\bf{1} ^TW=0$$ and according to 1st assumption for all such vectors we have:$$r_n^TVDVr_n=W^TDW\le 0$$which completes our proof.