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.

  • $\begingroup$ 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. $\endgroup$ – Brian Borchers Sep 18 '18 at 21:40
  • $\begingroup$ @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? $\endgroup$ – 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., $-A\in S_{+}^{n}$?


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