Show that the symmetric positive semidefinite cone $K = S_+^n$ is a proper cone.

Question

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.

Answer 1

The following fact might help in proving that the cone is pointed: A symmetric matrix $A$ belongs to $S_{+}^{n}$ if and only if all the eigenvalues of $A$ is positive. Now write $A = U \Lambda U^{T}$, where $\Lambda$ is a diagonal matrix containing all the eigenvalues of $A$. What can you say about the eigenvalues of $A$ if $A \in -S_{+}^{n}$, i.e., $-A\in S_{+}^{n}$?

Answer 2

1. 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!

2. $S_+^n$ is closed.

   Here is a link from Math stackexchange, proving $S_+^n$ is closed

3. $S_+^n$ has non-empty interior.

   We can use example $I_{n\times n}\in S_+^n$ and $I_{n\times n}\succ 0$