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.

  • 4
    $\begingroup$ 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. $\endgroup$
    – user98602
    Aug 1, 2016 at 3:31
  • $\begingroup$ @Mike Miller, don't we need to show that any matrix within some distance of A is also positive-definite? $\endgroup$ Aug 1, 2016 at 3:38
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    $\begingroup$ 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). $\endgroup$ Aug 1, 2016 at 3:44
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    $\begingroup$ 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. $\endgroup$ Aug 1, 2016 at 4:38
  • $\begingroup$ @MikeMiller why is it enough to check on a basis? $\endgroup$
    – peter
    Dec 21, 2018 at 21:43


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