Multiplicity of the eigenvalues of the sum of two matrices.

Question

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 ?

Answer 1

The conjecture is false. For example, take $$ A = \left[\matrix{ 1 & 0 & 0 & 0\\ 0 & 2 &0 & 0\\0 & 0 &3 & 0\\0 & 0& 0 & 4}\right]. $$ Then $A$ is symmetric, invertible, and has only simple eigenvalues, as requested.

Let $$ B = \left[\matrix{ -1 & 0 & 0 & 0\\ 0 & -2 &0 & 0\\0 & 0 &0 & 1\\0 & 0& 1 & 0}\right]. $$ Then $B$ is symmetric, and $A$ and $B$ do not commute, as requested.

Now $$ A + B = \left[\matrix{ 0 & 0 & 0 & 0\\ 0 & 0 &0 & 0\\0 & 0 &3 & 1\\0 & 0& 1 & 4}\right]. $$ Clearly $A+B$ has $0$ as an eigenvalue with multiplicity $2$, which shows that the conjecture is false.

Answer 2

This question ultimately was about the conditions for violation of the Wigner-von Neumann non-crossing rule, which was asked, and, answered here: mathoverflow.net/q/261826