Proof: $A=B^2$ for some positive definite symmetric $B$

Question

Let A be a real $ n \times n$ positive definite symmetric matrix.Show that $A=B^2$ for some positive definite symmetric matrix $B$.

So we can try and diagonalize $A$ such that $A=SJS^{-1}$then let $B=S^{-1}\sqrt{J}S $.

But what if one of the eigenvalues of $A$ is "missing" an eigenvector? Our $J$ and $B$ will no longer be positive definite, right?

This addresses an issue that the following answer: Square root of Positive Definite Matrix, does not address.