Let $f: \mathbb{C}^2 \to \mathbb{C}^n$ be the map $$ f(u, v) = (u^{n-1}, u^{n-2}v, \ldots, u^{n-k} v^{k-1}, \ldots, v^{n-1})^T$$ so that $f$ is the rational normal curve in homogeneous coordinates.
Note that the image of $f$ can be described as: $$ \operatorname{Im}(f) = \{ c (1, z, z^2, \ldots, z^{n-1})^T; c, z \in \mathbb{C} \} \cup \{ d (0, 0, \ldots, 0, 1)^T ; d \in \mathbb{C} \}. $$
Let $H$ be an hermitian $n \times n$ matrix such that $$ (v, Hv) > 0, $$ for any $0 \neq v \in \operatorname{Im}(f)$, where $(-, -)$ is the standard hermitian inner product on $\mathbb{C}^n$.
Are these conditions sufficient to guarantee that $H$ is positive definite? This question just popped in my head. Part of me thinks it would be too good to be true, in which case it should be easy to construct a counterexample, numerically. Another part of me thinks that the rational normal curve "twists" enough to guarantee that $H$ is positive definite.
I will think about it during the day, but I think it is an interesting question.
There is a counter example.
For $n=3\,,$ $$ f(u,v)=(u^2,uv,v^2)\,. $$ Let $H$ be the Hermitian matrix $$ H=\begin{pmatrix}1&0&0\\0&0&0\\0&0&1\end{pmatrix}\,. $$ which is only nonnegative semidefinite. Then, for $x\in{\rm Im}(f)\setminus\{0\}\,,$ $$ (x,Hx)=|u^2|^2+|v^2|^2>0\,. $$