Suppose we have a $n\times n$ symmetric positive semidefinite matrix $A$ which we can define it as a vector $v$ of $\frac{n^2+n}{2}$ elements, its upper triangular elements and diagonal elements. We have a ball of radius $\epsilon >0$ around $v$, is it true that in this ball we can find a vector such that its matrix is positive definite? I mean a positive semidefinite matrix with non-zero determinant.
I think it is true but I don't have any proof for it. If you give me some hint I would be grateful.
Best Answer
If $A$ is positive-semi definite we can diagonalize it as $A=U^* D U$ where $U$ is unitary and $D$ is a diagonal matrix with nonnegative diagonal entries. Its easy to pick a diagonal matrix $E$ so that $D+E$ has positive diagonal entries. Then $A+UEU^*$ is a positive definite matrix. Vectorizing $UEU^*$ as you did with $A$ gives you a vector which works. You can then scale the norm of $E$ to make the vector as small as you want.
A different way to think about this is, the set of positive semi definite matrices is the closure of the set of positive definite matrices. This means that if you put an epsilon-ball around any positive semi-definite matrix, it will contain positive definite matrices. Vectorizing can't meaningfully change your topology since we are in finite dimensions.