Why is the set of positive definite matrices in $\mathbb R^{n\times n}$ a positive cone

Question

The set of positive definite matrices in $\mathbb R^{n\times n}$ is geometrically a positive cone. This statement appears in almost every article on real positive definite matrices I read but without a proof. Where can I find a general proof, please? In addition, if I assume this statement is true, does it imply that the set of positive definite matrices is a manifold or a sub-manifold of $\mathbb R^n$ for some $n$?

Answer 1

It's immediate from the definitions. An $n\times n$ real matrix $A$ is positive definite if (1) it is symmetric ($A^t=A$) and (2) $v^tAv>0$ for all non-zero column vectors $v\in\mathbb R^n$. The 1st condition is linear, so defines a linear subspace of $\mathbb R^{n^2},$ of dimension $n(n+1)/2$ (easy exercise). The 2nd condition defines an open subset of $\mathbb R^{n(n+1)/2}$ which is a convex cone; ie a subset closed under linear combinations with positive coefficients. This is very easy to verify. To show that its open you use the fact that $(v,A)\mapsto v^tAv$ is a continuous function. Thus your set is an $n(n+1)/2$ dimensional manifold, a convex open subset of $\mathbb R^{n(n+1)/2}$.