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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.

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    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.2012-06-05

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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).