conditions for positivity invariance in inner products Let $R = R^{\top} \in \mathbb{R}^{m \times m}$ be a positive definite matrix. My question is: 
Under which conditions $u^{\top} R w > 0$ implies $u^{\top} w > 0$ ?
Rejecting the trivial case $R = I$.
Thanks in advance.

Assume there is an SPD $R$ such that for all $u$ and $v$ such that $u^TRw>0$ we have $u^Tw>0$. We show that this is not possible unless $R$ is a positive scalar matrix.
Let $e_i$ be the $i$th vector of the standard basis.
First we show that $R=(r_{ij})$ must be diagonal. Assume that $r_{ij}\neq 0$ for some $i$ and $j$. Note that $r_{ij}=e_i^TRe_j$. Take $u=e_i$ and $w=\mathrm{sign}(r_{ij})e_j$. Then $u^TRw=|r_{ij}|>0$, but $e_i^Te_j=0$.
Now we show that the diagonal elements of $R$ cannot be distinct. So assume that for some $i$ and $j$, $0<r_{jj}<r_{ii}$. Take $u=e_i-e_j$ and $w=e_i+e_j$, then
$$
u^TRw=(e_i-e_j)^TR(e_i+e_j)=r_{ii}-r_{jj}>0,
$$
($e_i^TRe_j=e_j^TRe_i=0$ since $R$ is diagonal) but
$$
u^Tw=e_i^Te_i-e_j^Te_i+e_i^Te_j-e_j^Te_j=0.
$$
Therefore
$$
R=\alpha I, \quad \alpha>0.
$$
P.S.: Actually you can avoid assuming that $R$ is SPD from the start. You get that the $\alpha$ at the end must be positive quite easily.
