We have a symmetric matrix $A$, with some entries specified and others not. We are trying to find the values of the unspecified entries so that the matrix $A$ becomes positive semidefinite. How can I prove that I can assume that the diagonal entries of $A$ are specified?
Intuitively, if we do not specify a diagonal entry, say $i$th entry, we can take it to infinity. Then the positive definiteness of $A$ is equal to the positive definiteness of the new matrix $A[-i,-i]$ where we remove the $i$th column and row. I do not know how to show this mathematically.
Best Answer
As a consequence of Gershgorin's Theorem, we know that the eigenvalues of a matrix $A$ live in balls $B(a_{ii},\sum_{j\neq i} |a_{ij}|)$ (centered at $a_{ii}$ of radius $\sum_{j\neq i} a_{ij}$). So if your matrix has real entries, but you have the freedom to choose the diagonal entries, then choosing each diagonal entry to be greater than the sum of the absolute values of the other entries in the same row will immediately imply that all of the eigenvalues of $A$ are positive, and therefore that $A$ is positive definite.