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$\mathcal M_n(\Bbb R)$ is endowed by some norm and let $\mathcal S_n(\Bbb R)$ the subspace of symmetric matrices. If we have $\mathcal O$ is an open set of $\mathcal S_n(\Bbb R)$ is it also an open set of $\mathcal M_n(\Bbb R)$? I ask this question because I proved that the set of symmetric definite positive matrices is open in $\mathcal S_n(\Bbb R)$ and I wonder if it is open in $\mathcal M_n(\Bbb R)$?

user326070
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2 Answers2

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A general fact of topology is that if a subspace $A \subseteq X$ is open, then any subset that is open in $A$ is also open in $X$. However, the space of symmetric matrices is not open in the space of matrices (except in the trivial case of $1 \times 1$ matrices). Small perturbations of symmetric matrices need not be symmetric; for instance, add any $\epsilon > 0$ to a single zero entry of the identity matrix to make it non-symmetric but still arbitrarily close to the identity. The same argument shows that the space of symmetric positive definite matrices is also not open.

Alex Provost
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  • This question say that the set of symmetric definite positive matrices is open!? – user326070 Mar 26 '16 at 15:57
  • @user326070 Inside the space of symmetric matrices, sure. Not inside the space of all matrices. Note that the top answer perturbs the diagonal elements, not a generic matrix element. The latter would make the resulting matrix non-symmetric. – Alex Provost Mar 26 '16 at 16:02
  • Sorry but see this question. It say also that the symmetric definite positive matrices is open in $\mathcal M_n(\Bbb R)$. Is it wrong? – user326070 Mar 26 '16 at 16:10
  • @user326070 Positive definiteness is an open condition. Symmetry is not. That question is about positive definite matrices, not necessarily symmetric ones. To recap: symmetric positive definite matrices are open in the space of symmetric matrices; positive definite matrices are open in the space of matrices. – Alex Provost Mar 26 '16 at 16:20
  • @user326070 Note that if you know that positive definite matrices are open in the space of all matrices, you get for free that symmetric positive definite matrices are open in the space of symmetric matrices. This is an immediate consequence of the definition of the subspace topology. – Alex Provost Mar 26 '16 at 16:28
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If the topology on $\mathcal{M}_n(\mathbb{R})$ is the same topology of $\mathbb{R}^{n^2}$, then note that the set of symmetric matrices will correspond to an hyperplane of $\mathbb{R}^{n^2}$ defined by the equations $a_{ij}=a_{ji}$. Thus, $X=\mathcal{S}_n(\mathbb{R})$ is open in $\mathcal{S}_n(\mathbb{R})$ but closed in $\mathcal{M}_n(\mathbb{R})$

Darío G
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