0
$\begingroup$

$\mathbb{S}_{+}^n$ is the set of positive semidefinite (and symmetric) real matrices of size $n\times n$. I have to prove that this set is a closed convex cone. How can I do?

$\endgroup$
  • 2
    $\begingroup$ Do you know the definitions of closed, convex, and cone? Can you show any of the three properties? $\endgroup$ – Mees de Vries Jan 16 at 13:49
  • $\begingroup$ It is not closed. ${ 1 \over n} I \to 0$. The other properties are just a matter of grinding through the definitions. $\endgroup$ – copper.hat Jan 16 at 15:05
  • $\begingroup$ I expect positive means positive definite as in $Ax\cdot x\geq0$ for all $x$. $\endgroup$ – SmileyCraft Jan 16 at 15:18
  • $\begingroup$ @SmileyCraft that would be positive semidefinite $\endgroup$ – LinAlg Jan 16 at 15:31
0
$\begingroup$

Note that $A \ge 0$ iff $x^TAx \ge 0$ for all $x$ iff $A \in \cap_x \{ B | x^T B x \ge 0\}$.

Note that for any $x$ that $\{ B | x^T B x \ge 0\}$ is a closed half space (hence convex). Since $0 \in \{ B | x^T B x \ge 0\}$ we see that it is a cone as well.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.