I'd like to know what is the convex cone generated by $\left\{ (h h^T, h) : h \in \Bbb R^{d\times1} \right\}$. It is known that $$\mathrm{cone} \left\{h h^T : h \in \Bbb R^{d \times1} \right\} = S_+^d$$ $$\mathrm{cone} \left\{h : h\in \Bbb R^{d \times1} \right\} = \Bbb R^{d \times1}$$
I am trying to prove or disprove that
$$ \left\{ (h h^T, h) : h \in \Bbb R^{d\times1} \right\} = S_+^d\times R^{d\times1} $$
Any idea or reference to the related literature would be appreciated.
Are you really interested in the convex cone, or in the convex envelop ? The latter is easily determined by taking the intersection of the half-spaces containing all the pairs $(hh^T,h)$. Their equations $\ell(M,h)={\rm cst}$ for some linear form $\ell$ over ${\bf Sym}_d\times R^d$, are actually $$(S^{\frac12}v)\cdot h\le\frac{|v|^2}4+{\rm Tr}(SM)$$ and are parametrized by pairs $(S,v)\in{\bf Sym}_d\times R^d$. Thus the convex envelop is $$\left\{(M,h)\in{\bf Sym}_d\times R^d|\forall(S,v)\in{\bf Sym}_d\times R^d,(S^{\frac12}v)\cdot h\le\frac{|v|^2}4+{\rm Tr}(SM)\right\}.$$ An elimination gives this envelop as the set $$\{(M,h)\in{\bf Sym}_d\times R^d|M\ge hh^T\}$$ where the inequality is the natural order (that of quadratic forms) in ${\bf Sym}_d$.
Edit. Now the cone is just the union of rays ${\mathbb R}_+(M,h)$ where $(M,h)$ run over the convex envelop. If $N$ is positive semi-definite and $k$ is a vector, then write $M=\frac1\mu N$ and $h=\frac1\mu h$. For $\mu>0$ large enough, one has $M\ge hh^T$ provided that $h\bot\ker N$. To summarize, the convex cone is made of pairs $(N,k)$ for which $N$ is positive definite or semi-definite, and $h\bot\ker N$. Hence the description of the cone: $$\{(N,k)\in S^d_+\times R^d|k\in R(N)\}$$
