I know that given two matrices $A$ and $B$, estimating the eigenvalues of $A + B$ by the eigenvalues of $A$ and $B$ is generally a non-easy problem. In particular, there are some results for matrices that commute (multiplicatively!), hermitian matrices etc.
In this case $B=\operatorname{diag}(1, 0,\dots,0)$ and the sum of the elements of every row of $A$ is $0$; each eigenvalue of $A$ is non-negative. I was wondering if the solution is known in this case, at least if one can say something about the sign of eigenvalues of $A+B$.
Thanks in advance.
Best Answer
The question is to know if the $0$ eigenvalue of $A$ can become negative when we add $B$. We can write specific results only if $B$ is a small perturbation of $A$; it is easier to assume that $A$ is fixed and $B=diag(x,\cdots,0)$ with a small positive $x$.
Let $U$ be the matrix obtained from $A$ deleting its first column and row. Then $\phi(\lambda)=\det(A+B-\lambda I)=\det(U-\lambda I)x+\det(A-\lambda I)$; $\phi(0)=\det(U)x$ and $\phi'(0)=-tr(adj(U))x-tr(adj(A))$. Since $0$ is a simple eigenvalue of $A$, $tr(adj(A))\not=0$ and $0$ does not burst in $2$ conjugate eigenvalues of $A+B$. We assume that $\det(U)\not=0$.
EDIT. Here we assume that $0$ is a simple eigenvalue of $A$ and (consequently) its other eigenvalues are $>0$. Then $tr(adj(A))>0$. Thus the conclusion is as follows:
If $\det(U)<0$, then $A$ admits a negative eigenvalue; otherwise $0$ gives birth to a positive eigenvalue.