Do the eigenvalues of $A+B$ have some relation with eigenvalues of $A, B$? I am wondering if I can say something about eigenvalues of $A+B$ in terms of eigenvalues of $A$, eigenvalues of $B$, under some conditions.  (In particular, $A$, $B$ of my interest are stochastic matrices, but it is okay for any general matrices.)
I know that there is no general rule for this, but I am wondering if we can say something nontrivial under a certain condition. I find one obvious case here Eigenvalues of product/sum of two matrices
but I want to say more. Any thought will be appreciated!
 A: For Hermitian matrices, as pointed out by podika in a comment under your question, there are some inequalities that relate the eigenvalues of $A+B$ to those of $A$ and $B$.
For general square matrices, however, apart from the obvious trace condition that $\sum_i\lambda_i(A)+\sum_i\lambda_i(B)=\sum_i\lambda_i(A+B)$, I don't think there is any relation between the eigenvalues of $A+B$ and those of $A$ and $B$. For instance, let
$$
A=\pmatrix{\lambda_1&a\\ 0&\lambda_2},
\,B=\pmatrix{\lambda_3&0\\ 1&\lambda_4},
\,A+B=\pmatrix{\lambda_1+\lambda_3&a\\ 1&\lambda_2+\lambda_4}.
$$
The characteristic polynomial of $A+B$ is then
$$
p(x)=x^2-\left(\sum_{i=1}^4\lambda_i\right)x
+(\lambda_1+\lambda_3)(\lambda_2+\lambda_4)-a.
$$
By adjusting $a$, we can fit $p$ to the characteristic polynomial of any matrix whose trace is $\sum_{i=1}^4\lambda_i$. In other words, given any six complex numbers $\lambda_1,\ldots,\lambda_6$ such that $\lambda_5+\lambda_6=\sum_{i=1}^4\lambda_i$, we can always construct two complex $2\times2$ matrices $A$ and $B$ such that the spectra of $A,B$ and $A+B$ are $\{\lambda_1,\lambda_2\},\{\lambda_3,\lambda_4\}$ and $\{\lambda_5,\lambda_6\}$ respectively.
