# Are the eigenvalues of the sum of matrices with real eigenvalues still real?

Say I have two matrices $$A$$ and $$B$$. I know that they each have real eigenvalues. Clearly if they're Hermitian, then their sum would also have real eigenvalues, since then $$A+B$$ is Hermitian. However, I'm wondering what happens if they are not necessarily Hermitian. Can we say anything about $$A+B$$? Are the eigenvalues of $$A+B$$ real? I have tried to come up with a counterexample but didn't manage and didn't get very far manipulating the characteristic equation. What if just one of them is Hermitian?

As an example, look at $$A=\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$$ and $$B=\begin{bmatrix} 0 & -1 \\ 0 & 0 \end{bmatrix}$$.
• (Never mind my previous comment) I think this also answers the question of Hermiticity by making $B_{21}=-1$ (thereby $B$ is Hermitian) and $A_{21}=2$. Thanks!
Every real square matrix $$C$$ is the sum of two real matrices $$A$$ and $$B$$ with real spectra, where $$A$$ can be chosen to be symmetric. In fact, let $$H$$ and $$K$$ be the symmetric and skew-symmetric parts of $$C$$, and let $$L$$ be the lower triangular part of $$K$$. Then one can take $$A=H+L+L^T$$ and $$B=-2L^T$$.
But obviously, not every real square matrix $$C$$ has a real spectrum.