Positive matrices are open An linear application $A:\mathbb R^n\to \mathbb R^n$ is positive when it is symmetric and besides that $\langle Ax,x\rangle\gt 0$ for every $x\neq 0$ in $\mathbb R^n$. I would like to prove the set of the positive linear applications is open in the set of symmetric applications.
I don't have any idea how to prove it.
How does this matrix look like? I think this question can be answered directly from that.
Any hint?
Thanks
 A: The operator norm of a matrix $A$ is $\|A\| = \sup_{\|x\| = 1} \|Ax\|$. The claim is that if $\|A - B\|$ is small, then $B$ is positive too. It suffices to prove that $\langle Bx,x \rangle > 0$ whenever $\|x\| = 1$.
Let $\alpha = \min_{\|x\| = 1} \langle Ax,x \rangle$. Since the unit sphere in $\mathbf R^n$ is compact we have $\alpha > 0$.
If $\|x\| = 1$ then
$$ \langle Ax , x \rangle - \langle Bx , x \rangle = \langle (A-B)x,x \rangle \le \|(A-B)x\| \|x\| \le \|A-B\|.$$ In particular, $\|x\| = 1$ implies
$$
\langle Bx , x \rangle \ge \alpha - \|A-B\|.$$
This is positive if $\|A-B\| < \alpha$.
A: Assume $A>0$. Then $A$ is invertible and $A\geq \|A^{-1}\|^{-1}\cdot I$. It follows that
$$
A+B\geq \|A^{-1}\|^{-1}\cdot I +B \geq(\|A^{-1}\|^{-1}-\|B\|)\cdot I>0
$$
whenever $\|B\|<\|A^{-1}\|^{-1}$. This shows that (in the set of symmetric, real matrices) the ball around $A$ with radius $\|A^{-1}\|^{-1}$ is contained in the set of positive matrices, namely let $C$ be a symmetric matrix with $\|A-C\|<\|A^{-1}\|^{-1}$, then $C=A+(C-A)$, which allows us to use the result above, with $B=(C-A)$.
Notation: For a description of the notation I use, check out the wiki page on positive definite matrices, where most of the properties I make use of are also mentioned.
Addendum: Why is A invertible, and how do we get the lower bound. Let $A>0$. Then $A$ is in particular self-adjoint, so the Finite Dimensional Spectral Theorem allows us to pick a basis $e_1,\ldots,e_n$ such that, with respect to this basis,
$$
A=\mbox{diag }(\lambda_1,\ldots,\lambda_n),
$$
where $\lambda_1\geq\lambda_2\geq\ldots\geq\lambda_n>0$ are the eigenvalues of $A$ in decreasing order. Now you can show that $\|A\|=\lambda_1$ and that $A$ is invertible, namely
$$
A^{-1}=\mbox{diag }(1/\lambda_1,\ldots,1/\lambda_n).
$$
In particular, you can show that
$$
\langle Ax,x\rangle\geq \lambda_n=1/(1/\lambda_n)=\|A^{-1}\|^{-1}
$$
for all $x$ with $\|x\|=1$, and that there is equality if $x=e_n$.
A: Is the following solution correct?
Let $\mathcal{A}$ be the set of positive linear applications. For each $x$ st $|x|=1$ define $ f_x: \mathcal{A} \to  \mathbb{R} $ by $\langle Ax,x\rangle$.
$f_x$ is continuous and $(0,\infty)$ is open. So
$f_x^{-1}((0,\infty))$ is Open. So,
$$
 \mathcal{A}=\bigcup_{|x|=1} f_x^{-1}((0,\infty))
 $$ is Open.
