SOLVED: Multiplicity of the eigenvalues of the sum of two matrices * Question Solved *
This question ultimately was about the conditions for violation of the Wigner-von Neumann non-crossing rule, which is still an open field of research. 
Thank you very much to all those who contributed. 
Edit in response to the comments received (see below for initial, disproved, conjecture):
Let $\boldsymbol{D} \in \mathrm{Diag}_n\left(\mathbb{R}\right) $ be an invertible diagonal matrix that has different, unique real values on the diagonal. 
Let $\boldsymbol{d} \in \mathrm{Diag}_m\left(\mathbb{R}\right) $ be a smaller diagonal matrix , with $m<n$. $\boldsymbol{d}$ is not necessarily invertible and could have repeated values on the diagonal. 
Finally, let $\boldsymbol{M} \in \mathrm{Mat}_{n,m}\left(\mathbb{R}\right)$ be a rectangular matrix that is full and random enough that it is of rank $m$ (i.e. $\boldsymbol{M}$ would be invertible where it not for $m \neq n $ ).
The conjecture is that $\boldsymbol{D} + \boldsymbol{M} \boldsymbol{d} \boldsymbol{D}^\top $ has simple eigenvalues (i.e. no multiplicities higher than 1). 
Thank you very much for any help towards demonstrating this result.
Initial Statement (disproved by counter-example in comments):
If $\boldsymbol{A}$ only has simple eigenvalues, are there properties concerning the multiplicity of the eigenvalues of $\boldsymbol{A} + \boldsymbol{B}$ ?
This question is much narrower than the general Additive Eigenvalue Problem conjectured by Horn (1962) and demonstrated in 1999 by Knutson and Tao. We do not ask what are the eigenvalues of $\boldsymbol{A} + \boldsymbol{B}$, but only whether they are simple. 
Here is the statement: 
Let $\boldsymbol{A}$ and $\boldsymbol{B}$ be two symmetric matrices (i.e. in $\mathcal{S}_n\left(\mathbb{R}\right) $). 
We suppose all eigenvalues of $\boldsymbol{A}$ are simple (i.e. multiplicity 1) and $\boldsymbol{A}$ is invertible. 
There are no restrictions on the eigenvalues of $\boldsymbol{B}$, in particular it is likely that $0$ is an eigenvalue of high multiplicity. 
$\boldsymbol{A}$ and $\boldsymbol{B}$ do not commute. 
I conjecture that the eigenvalues of $\boldsymbol{A} + \boldsymbol{B}$ are all simple (i.e. $\boldsymbol{A} + \boldsymbol{B}$ has, as $\boldsymbol{A}$, only eigenvalues of multiplicity 1). However, a proof is still elusive. Can anyone help ?
Thank you in advance.
 A: It's not true even if $A$ and $B$ have no common nontrivial invariant subspaces.
In dimension $3$, let $A$ be any symmetric matrix with  three distinct nonzero eigenvalues and eigenvectors which have all their entries nonzero, and
$$B = \pmatrix{0 & 0 & 0\cr 0 & 0 & 0\cr 0 & 0 & 1\cr} - A$$
Note that $A$ and $A+B$ have no nontrivial invariant subspaces in common, so neither do $A$ and $B$.
A: While your conjecture is wrong, it still is quite unlikely that $A+B$ has a double eigenvalue. If $A$ and $B$ are symmetric, the set of matrices $B$ such that $A+B$ has a double eigenvalue has codimension 2, not 1. So a crossing is "doubly more unlikely" than what one would expect.
A striking manifestation of this phenomenon is the so-called avoidance of crossing: take two random symmetric matrices $A,B$, and plot the eigenvalues of $A+tB$ as a function of $t$. Generically (i.e., "with probability 1"), this plot contains no crossings; and it often contains lines that draw very close, seem to meet and then suddenly diverge; for instance, see the plots in this web page.
This phenomenon is discussed in more detail in Chapter 10 of Lax's Linear algebra and its applications.
