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

I'm not sure if the set of symmetric positive-definite matrices is open in the set of symmetric matrices. I'm almost certain that it is not open in the larger set of n by n matrices. So far most of the proofs I've seen don't use symmetry, so I'm not sure if they are correct. Any hints? I've been trying to use that the set of positive-definite matrices is path connected, but I can't come up with a solution. Thanks in advance.

• What does positive-definite mean if you're not a symmetric matrix? Anyway, yes - this is equivalent to saying that $x^T A x$ is positive for all nonzero $x$, which is equivalent to saying that $x_i^T Ax_i$ is positive for the basis vectors $x_i$. This last condition, then, is clearly an open condition. – user98602 Aug 1 '16 at 3:31
• @Mike Miller, don't we need to show that any matrix within some distance of A is also positive-definite? – Saul Alinsky Aug 1 '16 at 3:38
• Yes, Sylvester's criterion implies that any matrix $A'$ which is sufficiently close to a positive-definite matrix $A$ is positive-definite (of course, provided $A'$ is symmetric). – Alex Ravsky Aug 1 '16 at 3:44
• Here's a more direct proof. Sylvester's criterion mentioned by @AlexRavsky shows that the set of positive-definite matrices is the intersection of the set of symmetric matrices and the set $\{A\mid\text{det}A_k>0,k=1,2,\ldots,n\}$, which is the intersection of inverse images of a open set under the continuous determinant map. – Cave Johnson Aug 1 '16 at 4:38
• @MikeMiller why is it enough to check on a basis? – peter Dec 21 '18 at 21:43