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

Solution 1:

Let $A$ be a symmetric positive definite matrix, and let $$ \delta = \inf x\cdot Ax > 0, $$ where the infimum is taken over all unit vectors $x$.

Now let $A+B$ be a symmetric perturbation of $A$. Then for any unit vector $x$, we have $$ x\cdot(A+B)x = x \cdot Ax + x \cdot Bx, $$ and $$ |x \cdot Bx|\leq\|B\|\|x\|^2 = \|B\|, $$ so that $$ x \cdot(A+B)x \geq \delta - \|B\|. $$ This shows that if $B$ is small enough then $A+B$ is positive definite.

Note that the set of symmetric positive definite matrices is not open in the space of all matrices (because a general perturbation destroys symmetricity), but open within the subspace of symmetric matrices, with respect to the subspace topology.

Note also that this argument works verbatim for the general positive definite case with general perturbations.