Relation between the eigenvalues of symmetric $A$, $B$ and $A+B$ when $AB=(BA)^{T}$

Question

Let $A$ and $B$ be two real symmetric matrices. If $AB=(BA)^{T}$ then, is there any relation between the eigenvalues of $A+B$ and eigenvalues of $A$, $ B$?

If $A$ and $B$ be two real symmetric matrices such that $AB=BA$, then the eigenvalues of $A+B$ is the sum of eigenvalues of $A$ and $B$ in some order. The similar way, can we express the eigenvalues $A+B$ using the eigenvalues of $A$ and $B$ when $AB$=transpose of $BA$.

Answer 1

Note that for any two symmetric matrices $A, B$, we have $$ (BA)^T = A^TB^T = AB $$ so assuming that $(BA)^T=AB$ doesn't actually let us conclude anything that we couldn't already conclude from $A$ and $B$ being symmetric.

Answer 2

If A and B has same eigenvector, then $Ae=\lambda e$ and $Be=\mu e$ $\implies$ $(A+B)e=(\lambda+\mu)e$.

If there is no relationship between eigenvectors, then there cannot be any relationship between the eigenvalues. However, if B is a perturbation of A, then the eigenvectors might be similar, in which case the above statement holds approximately.

Also, see this.