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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.

Hang
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  • When you say $B$ is in a small neighborhood of $A$, do you mean that $|B-A| < \epsilon$ (operator norm), $|B-A|_F < \epsilon$ (Frobenius norm), or some other norm? – JimmyK4542 Dec 17 '18 at 01:41
  • I just mean the topology of $\mathbb R^{n^2}$, but I think I am also interested in the question for different topology. – Hang Dec 17 '18 at 01:49
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    @Hang You can find some discussion on the topic here https://math.stackexchange.com/questions/226486/the-openness-of-the-set-of-positive-definite-square-matrices – AnyAD Dec 17 '18 at 02:07
  • You can always find a PDS diagonal matrix as close as you want to the zero matrix. Does this help? – Wintermute Dec 18 '18 at 17:49
  • @Wintermute What does PDS stand for? – Hang Dec 18 '18 at 17:52
  • @Hang PDS = positive definite symmetric – Wintermute Dec 18 '18 at 17:54

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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 $B>\epsilon A$.