We say a cone at the origin in $R^n$ means that it is an intersection of finitely many halfspaces, i.e. 
$$C=\bigcap\_{i\in I}H\_i,\text{ where }|I|<\infty.$$


A cone is strongly convex if $C\cap -C=\{0\}$. Then we can define the linear dimension of $C$ as 
$$ldim(C):=dim(C\cap -C).$$


Here we assume the dimension of $C$ is always $n$. (NOTE! not $ldim(C)=n$ but $dim(C)=n$.) If $ldim(C)=k>0$, then can we use only $n-k$ halfspaces to represent the original cone $C$ as 
$$C=\bigcap\_{i\in [n-k]}H\_i$$


Of course for $(n,k)=(a,a)$ or $(a,a-1)$ or $(a,a-2)$ such cases can be reduced into $R^2$ or $R^3$ is obvious. Here $(n,k)$ means it is a $C$ in $R^n$ with whose $dim(C)=n,ldim(C)=k$. But I don't know how to prove it for general situations. 



The minimal number of halfspaces to represent a convex but non strongly convex cone
This is false in the other cases.


Let $\alpha=2\pi/n$ and observe that the interior of the polygon with $n$ given by $(cos(i\alpha),sin(i\alpha))\_{i=0,\dots,n}$ is given by $n$ affine inequalities (of degree $1$ but with constants), but not less.


Hence, the cone in $\mathbb{R}^3$ generated by $(cos(i\alpha),sin(i\alpha),1)\_{i=0,\dots,n}$ is not given by less than $n$ linear inequalities, and is thus not the intersection of less than $n$ halfspaces (just consider the intersection of your halfspaces with the plane where the third coordinate is $1$).


The same argument works for the cone in $\mathbb{R}^4$ generated by $(cos(i\alpha),sin(i\alpha),1,0)\_{i=0,\dots,n}$ and $(0,0,0,1)$ and $(0,0,0,-1)$ (intersect with the plane where the third coordinate is $1$ and the last is $0$). 


