# Set of symmetric positive semidefinite matrices is a full dimensional convex cone.

If $S_n^+$ is the set of all symmetric positive semidefinite $n \times n$ matrices with entries in $\mathbb{R}$, how does it follow that it is a full dimensional closed convex cone in $\mathbb{R}^{n^2}$?

I don't understand the relation between positive semidefiniteness and convexity of a cone.

-
It doesn't just follow, you have to do some work. What do you mean by 'full dimensional'? –  copper.hat Jun 3 '12 at 23:35
Hello, I got the proofs for convexity and the fact that it is a cone but I don't understand the proof for closedness. Also full dimensional means that the dimension of the cone is n^2. But, is that correct? Shouldn't the dimension of the cone be n(n+1)/2? –  nada Jun 5 '12 at 18:58
I gave a proof of closed below. Note that $S_n^+ = f_1^{-1} \{ 0 \} \cap f_2^{-1} [0,\infty)$ using the notation below. The dimension can be no larger than the dimension of the subspace of symmetric matrices, which is $\frac{1}{2} n (n+1)$. –  copper.hat Jun 5 '12 at 22:29
Okay, thank you. Also, what is the relative interior of Sn+ ? As far as I can think, it hould be the convex cone of positive definite symmetric matrices, but could you help me out with the reasoning please? Is it also closed? –  nada Jun 5 '12 at 22:36
Well, that is another question. You need to show that $\mathbb{aff} S_n^+$ is the set of symmetric matrices. Then show that if $A\geq0$ but not $A>0$, then there is a symmetric matric that is not positive semi-definite 'nearby'. Also show that if $A>0$, then all symmetric matrices 'nearby' are also positive definite. This shows that $\mathbb{ri} S_n^+$ is the set of positive definite matrices. –  copper.hat Jun 5 '12 at 22:57

For closed, note that the functions $f_1:\mathbb{R}^{n\times n} \to \mathbb{R}^{n\times n}$ given by $f_1(A) = A-A^T$, and $f_2: \mathbb{R}^{n\times n} \to \mathbb{R}$ given by $f_2(A) = \min_{||x||=1} \langle x, A x \rangle$ are continuous, so the sets $f_1^{-1} \{ 0 \}$ and $f_2^{-1} [0,\infty)$ are closed.
For convex, it is straightforward to check that if $\lambda \in [0,1]$, and $A,B \in S_n^+$, then $\lambda A + (1-\lambda)B \in S_n^+$ (just use the definition of positive semi-definiteness).
For cone, again it is straightforward to check that if $t\geq 0$ (0r whatever your definition of cone allows), then when $A \in S_n^+$, then $t A \in S_n^+$ (again, just use the definition).