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Let $S$ be the set of all symmetric positive definite matrices of size $n\times n$. Which of the following statements are true?

(a) $S$ is closed in $\mathbb{M}_n(\mathbb{R})$.
(b) $S$ is connected in $\mathbb{M}_n(\mathbb{R})$.
(c) $S$ is compact in $\mathbb{M}_n(\mathbb{R})$.

Only the option (a) & (b) are right. I guess that it is not bounded so (c) is not true. Am I correct?

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Can you prove it is not bounded? Do a particular case with $\,n=2\,$ and check you can easily construct a symmetric pos. def. matrix with as high a norm as wanted (I'm guessing you're taking the euclidean one in $\,\Bbb M_n(\Bbb R)\leq \Bbb R^{n^2}\,$ ...) – DonAntonio Jan 19 '13 at 17:39
a is false @etuku – learnmore Nov 18 '15 at 11:53

It is just connected infact it is path connected. for $A,B$ such matrices we have $x^TAx\ge 0$, $x^TBx\ge 0$ so for $\lambda \in [0,1]$ we get $x^T[\lambda A+(1-\lambda)B]x\ge 0$.

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can you confirm me that is it closed or not? i think it is closed not fully sure. – etuku Jan 19 '13 at 17:48
It is the inverse image of $0$ by the map $f(A) \mapsto (a_{i,j} - a_{j,i})_{i\neq j}$ which is clearly continuous. – Alexander Thumm Jan 19 '13 at 17:50
thanks.then it is closed. – etuku Jan 19 '13 at 17:50
I must have somehow read only half of the question. The set of positive definite symmetric Matrices is not closed. To see this, $\lambda I$ is in $S$ for $\lambda > 0$ but not for $\lambda = 0$. – Alexander Thumm Jan 19 '13 at 17:54

As La Belle Noiseuse said, $S$ is a convex set; $S$ is also open in a vector space of dimension $n(n+1)/2$; thus $S$ is homeomorphic to the open ball of $\mathbb{R}^{n(n+1)/2}$ for an arbitrary norm.

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