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I have read that the interior of $S^n_+$ in $S^n$ (real symmetric matrices of size $n\times n$, $n \in \mathbb{N}^*)$ is $S^n_{++}$. I am trying to prove it.

I have that there is no open set bigger than $S^n_{++}$ contained in $S^n_{+}$: if there were such a set, call it $V$, then taking $A \in V \setminus S^n_{++}$ and $\epsilon > 0$ small enough, $A - \epsilon I_n$ would be in $V$ (by openness), hence in $S^n_{+}$ but not in $S^n_{++}$.

But $0$ being an eigenvalue of $A$, $-\epsilon$ would be an eigenvalue of $A - \epsilon I_n$, absurd because this matrix is in $S^n_{+}$ hence has all eigenvalues $\geq 0$.

However I am struggling to prove that $S^n_{++}$ is open. I am tempted to say that $A\mapsto Sp(A)$ is continuous, because $A \mapsto \chi_A$ is continuous (the coefficients of $\chi_A$ are polynomials in coefficents of $A$) and roots of polynomials are continuous wrt the coefficents of the polynomial.

However i have also read that the interior of $S^n_{+}$ in $\mathcal{M}_n(\mathbb{R})$ is empty, and I don't see what goes wrong in my proof in this case, so I guess there is a mistake. Can someone pinpoint it and provide a correct proof, hopefully in the same spirit?

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    $\begingroup$ For any $A \in S^n_+$ and nonsymmetric matrix $B \in \mathcal{M}_n(\Bbb R)$, $A + \epsilon B$ is a nonsymmetric matrix in $\mathcal{M}_n(\Bbb R)$ for arbitrarily small $\epsilon > 0$, so every neighborhood of $A$ in $\mathcal{M}_n(\Bbb R)$ contains elements not in $S^n_+$, that is, $A$ is not in the interior of $S^n_+$ in $\mathcal{M}_n(\Bbb R)$. $\endgroup$ – Travis Sep 8 '16 at 7:07
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Your argument correctly allows us to deduce that $S^{n}_{++}$ is an open set relative to the set of symmetric matrices. Another argument that works is to note that $A \mapsto \det(A)$ is continuous.

Of course, $S^n_{++}$ is not open relative to the larger space $\mathcal M_n$. In particular, the symmetric matrices form a proper subspace of $\mathcal M_n$, and it follows that symmetric matrices has an empty interior within $\mathcal M_n$, as does any subset of the symmetric matrices.

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  • $\begingroup$ Thanks, but my question is rather: "Where does my reasoning (apparently correct in $S^n$ fail when applied in $M_n$?" $\endgroup$ – P. Camilleri Sep 8 '16 at 13:52
  • $\begingroup$ Oh, that's easy: eigenvalues alone are insufficient to determine that a matrix is symmetric. By the usual definition, positive semidefinite matrices are necessarily symmetric. Your argument could be directly applied to $\mathcal M_n$ show, for instance, that the set of matrices whose eigenvalues all have positive real part is open. $\endgroup$ – Omnomnomnom Sep 8 '16 at 13:54
  • $\begingroup$ Oh yes, I was too focused on eigenvalues. Thanks! $\endgroup$ – P. Camilleri Sep 8 '16 at 14:23

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