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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?

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As an example, look at $A=\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$ and $B=\begin{bmatrix} 0 & -1 \\ 0 & 0 \end{bmatrix}$.

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  • $\begingroup$ (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! $\endgroup$
    – tmph
    Jan 19 at 16:28
  • $\begingroup$ @tmph Yes that's a nice way to make one of them be Hermitian. $\endgroup$
    – Ian
    Jan 19 at 16:29
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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.

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