$$C = conv\{xx^T | x\succeq0\}$$

As far as I know, $x\succeq0$ means x is a symmetric, positive semidefinite matrix, therefore the $xx^T$ here is also symmetric and positive semidefinite.

As the "Positive Semedefinite Cone" is defined as $\mathbb{S}^{n}_{+} = \{\mathbf{X}\in\mathbb{S}^{n}: \mathbf{X}\succeq\mathbf{0}\}$, then $\{xx^T | x\succeq0\}$ should've already been a convex cone here.

How to understand this $C$, does it over-define a convex set?

Thank you!

  • $\begingroup$ I'm not sure what the question is... what property of $C$ do you want to understand? $\endgroup$ – user7530 Jan 16 '18 at 8:43
  • $\begingroup$ sorry for the vagueness, I just added some explanation $\endgroup$ – XJY95 Jan 16 '18 at 8:53
  • $\begingroup$ You've written $xx^{T}$ with a lower case $x$, which would normally refer to a vector $x$ rather than a matrix. $x \succeq 0$ could be an odd way of saying that the elements of the vector $x$ are nonnegative. I wonder if you're misreading your source. It would help if you could link to it. $\endgroup$ – Brian Borchers Jan 17 '18 at 5:22
  • $\begingroup$ sorry I missed this one, x here is actually a vector $\endgroup$ – XJY95 Jan 22 '18 at 3:08

I don't know what "over-define" means. Certainly each non-extreme element of any convex set $C$ has more than one representation as a convex combination of elements of $C$, so it is over-defined in one sense. I also don't completely understand your definition of $C$.

The set $D=\operatorname{conv}\{XX^T: X\in \mathbb S_+^n\}$ is equal to the set $\mathbb S_+^n$ as follows. Among the elements of $\mathbb S_+^n$ are the rank $1$ matrices of form $X=vv^T$ for $v\in\mathbb R^n$. But the convex hull of such matrices is the set of psd matrices, so $D \subseteq \mathbb S_+^n$. Since elements of $\mathbb S_+^n$ posessess psd square roots, each element $Y\in\mathbb S_+^n$ can be written in the form $Y=XX=XX^T$ for some $X\in\mathbb S_+^n$. Hence $\mathbb S_+^n\subseteq D$. Hence $D=\mathbb S_+^n$.

If your $C$ is my $D$ then it true that your definition of $C$ is another equivalent characterization of the psd matrices.

  • $\begingroup$ Thank you! It's kind of wired that I was told the x here is just a vector... $\endgroup$ – XJY95 Jan 17 '18 at 8:39

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