Let $A\in \mathbb{R^{n\times n}}$ be a symmetric negative difinite matrix and
$D\in \mathbb{R}^{n\times n}$ be a diagonal matrix $D = \mathrm{diag}\{d\_i\}, (d\_i < 0)$.
From Weyl's inequality, the maximum eigenvalue of the sum of these matrices $S = A + D$ can be evaluated as follows.


$\sigma\_{1} \leq \alpha\_{1} + \delta\_{1}$,


where $\sigma\_1\geq \cdots \geq \sigma\_n$, $\alpha\_1\geq \cdots \geq \alpha\_n$ and $\delta\_1\geq \cdots \geq \delta\_n$, are eigenvalues of $S$, $A$ and $D$ respectively.


Weyl's inequality can be applied to the sum of two Hermitian matrices. However here we have a stricter condition for $A$ and $D$. Therefore, I am considering the possibility of obtaining a smaller upper bound of $\sigma\_{1}$.
I would appreciate it if you could give me some advice.



Change in the largest eigenvalue due to perturbation of diagonal components of a symmetric matrix
You don't really have any stricter conditions, and in fact it is simple to reduce the general case to yours: every matrix is diagonal in *some* basis, and every matrix is negative definite if you subtract a suitable multiple of the identity to it.


So no, there can be no better bounds than the general case.


