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

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.

• As you've written, you need to show that $K$ is closed, pointed, and has a non-empty interior (not just that the cone is non-empty) Please show what you've attempted to do towards proving those things. – Brian Borchers Sep 18 '18 at 21:40
• @BrianBorchers I have updated my question on how to prove its convexity. I can see the closedness topologically but I can't seem to do it algebraically. For pointed, I can see that it ultimately boils down to showing (Sn+)∩(−Sn+) ={0}. But I can't seem to show it as well. Any guidance on these part? – Weiting Chen Sep 18 '18 at 23:00

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., $$-A\in S_{+}^{n}$$?