# What is the convex cone generated by the pair of rank 1 matrix and its eigenvector?

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.

• Do you agree with my edits? Jun 5, 2022 at 13:26

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: $$\{(N,k)\in S^d_+\times R^d|k\in R(N)\}$$