Suppose $A=(a_{ij})$ is a (symmetric) positive-definite matrix, and $B$ is another symmetric matrix.
Question: If $B$ is in a small neighborhood $U$ of $A$, then it seems that $B$ should also be positive-definite. Moreover for what value of $\epsilon>0$ we can find a neighborhood $U$ so that $B>\epsilon A$?
If both $A$ and $B$ are diagonal matrices, then this is trivial. But in general since we can only diagonalize one of them and I am afraid there will be some issue.
If you consider $\mathbb{R}^{n^2}$, then that does not work because $B$ may be non-symmetric. Thus we assume that $B=A+H$ where the small matrix $H$ goes through the vector space of symmetric matrices.
Let $spectrum(A)=\{\lambda_1\leq\cdots \leq \lambda_n\}$ (note that $\lambda_1>0$). Let $\epsilon\in[0,1)$ and $\alpha >0$ (to be determined). Since the norms are equivalent, we'll use the spectral norm defined, on the symmetric matrices, by $||S||=\rho(S)$ (the spectral radius of $S$).
We assume that $||H||<\alpha$, that implies $x^TBx\geq x^TAx-\alpha$ when $||x||_2^2=1$ (that we suppose in the sequel).
We choose $\alpha<\lambda_1$. Then $x^TBx\geq \lambda_1-\alpha >0$ and $B$ is symmetric$>0$.
Now $x^T(B-\epsilon A)x\geq x^TAx(1-\epsilon)-\alpha\geq \lambda_1(1-\epsilon)-\alpha$.
We choose $\alpha<\lambda_1(1-\epsilon)$. Then $B>\epsilon A$.