Is the set of positive-definite symmetric matrices open in the set of all matrices?

Question

So if $M_{n\times n}(\Bbb R)$ is the set of square real matrix of order $n$ then it is well know that the position $$ \langle A,B\rangle:=\text{tr}(AB^T) $$ defines an inner product on $M_{n\times n}(\Bbb R)$ so that this set is a topological vector space.

Now I'd like to prove that the subset $$ \text{Sym}_{n\times n}^+(\Bbb R):=\big\{A\in M_{n\times n}(\Bbb R):A=A^T\wedge \vec v\cdot A\vec v>0\,\,\text{for any }\vec v\in\Bbb R^{n}\setminus\{0\}\big\} $$ is open. So first of all I observed that if $\text{Sym}_{n\times n}(\Bbb R)$ and $M^+_{n\times n}(\Bbb R)$ and are the sets of symmetric matrices and the set of positive definite matrices, that is $$ \text{Sym}_{n\times n}(\Bbb R):=\big\{A\in M_{n\times n}(\Bbb R):A=A^T\big\}\\ M^+_{n\times n}(\Bbb R):=\big\{A\in M_{n\times n}(\Bbb R):\vec v\cdot A\vec v>0\,\,\text{for any }\vec v\in \Bbb R^n\setminus\{0\}\big\} $$ then $$ \text{Sym}_{n\times n}^+(\Bbb R)=\text{Sym}_{n\times n}(\Bbb R)\cap M_{n\times n}^+(\Bbb R) $$ so that the result would follow immediately showing that $M_{n\times n}^+(\Bbb R)$ is an open set disjoint form the boundary of $\text{Sym}_{n\times n}(\Bbb R)$ because this last is a linear subspace and any linear subspace of a finite dimensional t.v.s is closed so that in this case the above intersection is exactly equal to the intersection $$ \operatorname{int}\Big(\text{Sym}_{n\times n}(\Bbb R)\Big)\cap M_{n\times n}^+(\Bbb R) $$

Now let be $\varphi:M_{n\times n}(\Bbb R)\rightarrow\Bbb R^{n\cdot n}$ the function defined through the equation $$ \varphi(X):=(X_1,\dots ,X_n)=\big((x_{1,1},\dots,x_{1,n}),\dots,(x_{n,1},\dots,x_{n,n})\big) $$ for any $X\in M_{n\times n}(\Bbb R)$ and thus we observe that it is an isomorphism (moreover it is not hard to show that it is an isometry) and thus a homeomorphism becuase any ismorphism between finite dimensional t.v.s is an homeomorphism. So the statement can be proved showing that $\varphi\Big[M_{n\times n}^+(\Bbb R)\Big]$ is open and disjoint form the boundary of $\varphi\Big[\text{Sym}_{3\times 3}(\Bbb R)\Big]$ but unfortunately I was not able to prove this; alternatively the statement can be proved aslo showing that $\varphi\Big[\text{Sym}_{n\times n}^+(\Bbb R)\Big]$ but I yet was not be able to do this.

Finally I have to point out that here I found a question where is exatly ask to prove what I ask now but unfortunately it seems it is only proved that the set of positive matrices is open: indeed no answers there was accepted. Anyway I would not mind now that even here it proves that the set of positive matrices is open: indeed in the Ovi's answer there is something is not clear.

So could someone help me, please?

Answer 1

Let $A$ be a symmetric positive definite matrix, and let $$ \delta = \inf x\cdot Ax > 0, $$ where the infimum is taken over all unit vectors $x$.

Now let $A+B$ be a symmetric perturbation of $A$. Then for any unit vector $x$, we have $$ x\cdot(A+B)x = x \cdot Ax + x \cdot Bx, $$ and $$ |x \cdot Bx|\leq\|B\|\|x\|^2 = \|B\|, $$ so that $$ x \cdot(A+B)x \geq \delta - \|B\|. $$ This shows that if $B$ is small enough then $A+B$ is positive definite.

Note that the set of symmetric positive definite matrices is not open in the space of all matrices (because a general perturbation destroys symmetricity), but open within the subspace of symmetric matrices, with respect to the subspace topology.

Note also that this argument works verbatim for the general positive definite case with general perturbations.