Say I have a matrix $A \in R^{p \times p}$ which is symmetric and with non-negative diagonal entries (i.e. $a_{ii} \geq 0 \forall i\in \{1, \ldots, p\}$). However, $A$ is not positive (semi-)definite.
I want to somehow "perturb" the diagonal elements of $A$ so that it becomes positive (semi-)definite. The obvious way would be to just add $\alpha I$ where $I$ is the identity matrix and $\alpha \geq -\min_i \{\lambda_i(A)\}$, with $\lambda_i(A)$ being the $i$th eigenvalue of $A$. However, the diagonal elements of $A$ are on a different scale so I would rather add a different constant on the diagonal entries; that is I'm looking for a value $\alpha > 0$ such that $A + \alpha B$ is positive (semi-)definite for any $B := \text{diag}(b_1, \ldots, b_p)$ with $b_i > 0$. Can I always find an $\alpha$ value for any such matrix $B$ to ensure $A +\alpha B$ is positive (semi-)definite and what would the lowest value of $\alpha$ have to be?
Since the matrix is symmetric all eigenvalues are real. Further, there is a known result called Grishgorin theorem which states that eigenvalues are in circles with center at $a_{jj}$ and radius $\sum_{k} |a_{jk}|$. So, the maximum of radiuses of Grishgorin circle i.e. max of the sums above will make it.