# set of all symmetric non-negative definite matrices are closed or not

Can anyone tell me please that set of all symmetric non-negative definite matrices are closed or not in $\mathbb{M}_n(\mathbb{R})$ with usual topology

-

• If $S$ is a matrix and $x\in\Bbb R^n$, the map $f_x\colon S\mapsto x^tSx$ is linear hence continuous.
• Let $S^+_n(\Bbb R)$ the set of symmetric non-negative definite matrices. Then $$S^+_n(\Bbb R)=\bigcap_{x\in\Bbb R^n}f_x^{-1}([0,+\infty)).$$
• As $f_x^{-1}([0,+\infty))$ is closed in $\mathcal M_n(\Bbb R)$, $S_n^+(\Bbb R)$ is closed as an arbitrary intersection of such sets.
-
I am confused by the answer of you and the link given by panu.please guide me properly –  khuku Jan 20 '13 at 14:39
So, what don't you understand? –  Davide Giraudo Jan 20 '13 at 14:39
they give different answer.there says that it is not closed –  khuku Jan 20 '13 at 14:43
Positive definite is not the same thing as non-negative definite (which means $x^tSx\geqslant 0$ for all $x$). –  Davide Giraudo Jan 20 '13 at 14:48
I can't understand how $S^+_n(\Bbb R)=\bigcap_{x\in\Bbb R^n}f_x^{-1}([0,+\infty)).$? Please explain. –  S.Panja-1729 Feb 14 at 19:08