**\* 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.



SOLVED: Multiplicity of the eigenvalues of the sum of two matrices
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](https://www.uam.es/personal_pdi/ciencias/fchamizo/oscuro/avoidance.html).


This phenomenon is discussed in more detail in Chapter 10 of Lax's *Linear algebra and its applications*.


