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.