I am reading on Convex Optimization by Stephen Boyd.
I want to show that the symmetric positive semidefinite cone $K = S_+^n$, where $S_+^n$ is a set of symmetric $ n \times n$ matrices, is a proper cone
i am using the definition that a cone $K \subseteq \mathbb{R}^{n}$ is called a proper cone if
K is convex, closed, pointed (contains no line) and nonempty.
Looking through, $S_+^n$ is indeed a convex cone, but I can't seem to prove to myself that it is also indeed a proper cone.
This is what I know so far:
To prove $S_+^n$ is convex, we can prove it by using:
$X \in S_+^n \iff Z^TXZ \geq 0, \forall Z$
$X \geq 0$, $Y \geq 0$
$Z^T(\theta_1 X + (\theta_2 Y)Z$
$ = \theta_1Z^TXZ + \theta_2Z^TYZ$
Hence, since $X$ and $Y$ are affine, they are convex and hence, $K$ is convex.
Best Answer
Pointed (contains no line.)
We prove this by contradiction. If $K=S_+^n$ contains a line. Then there exists $A,B\in S_+^n$ and $A \ne B$ such that for any $\theta\in \mathbb{R}$, $\theta A+(1-\theta)B\in S_+^n$. Now if there exists $x\in \mathbb{R}^n$ s.t. $x^TAx\ne x^TBx$, without loss of generality. we assume $x^TAx > x^TBx$ then we can find $\theta$ s.t. $x^T(\theta A+(1-\theta)B)x<0$.
\begin{align} x^T(\theta A+(1-\theta)B)x = \theta x^T Ax+(1-\theta) x^TBx< 0 \end{align} if and only if $$\theta(x^TAx-x^TBx)<-x^TBx\Leftrightarrow \theta <-\frac{x^TBx}{x^TAx-x^TBx} $$ which implies $\theta A+(1-\theta)B\not \in S_+^n$. So there doesn't exists $x$ s.t. $x^TAx=x^TBx$ which $A=B$ is the case. Contradiction!
$S_+^n$ is closed.
Here is a link from Math stackexchange, proving $S_+^n$ is closed
$S_+^n$ has non-empty interior.
We can use example $I_{n\times n}\in S_+^n$ and $I_{n\times n}\succ 0$