TITLE: Maximizing the Smallest Eigenvalue of a Diagonally Dominant Matrix
QUESTION [10 upvotes]: Assume that we have a full-rank diagonally dominant matrix $A$, all the diagonal elements of which are positive, all the non-diagonal elements are negative, and the sum of the absolute values of the non-diagonal elements is equal to the diagonal element. More precisely:
\begin{equation}
A=[a_{i,j}] \qquad a_{i,i}>0 \qquad a_{i,j} \leq 0 \textrm{ for $i \neq j$} \quad \textrm{and } \quad a_{i,i}=\sum _{j\neq i}|a_i,_j |
\end{equation}
We also have a positive diagonal matrix $D$ whose trace is constant and equal to $K$:
\begin{equation}
d_{ij}=0 \textrm{ for } i \neq j \qquad d_{ii} \geq 0 \quad \textrm{and } \quad  K=\sum_{i}d_{i,i}
\end{equation}
What is the matrix $D$ such that the smallest eigenvalue of the matrix sum $T=A+D$ is as large as possible?
In other words, how can we find the $d_{ii}$ 's such that we maximize the smallest eigenvalue of $T=A+D$?
Thank you all in advance! :-)

REPLY [3 votes]: If the A matrix were symmetric (so the eigenvalues are real), then you could just solve a semidefinite programming problem (SDP) to find the matrix D (and 'lambda'). In particular, maximizing the smallest eigenvalue ('lambda') of the matrix A + D in your case would be equivalent to the SDP (over variables D and lambda):
max   lambda
such that:  A + D >= lambda I
            Trace(D) = K
            D_{ii} >= 0
where the first '>=' denotes 'greater-or-equal in the cone of positive semidefinite matrices', and I is the identity matrix. 
There are several MATLAB-based packages in which you can formulate and solve problems like this (for instance, Yalmip and CVX). You'll probably also need a solver for SDPs (e.g., sedumi or sdpt3). 
Good luck,
  -Dan