Real part of the eigenvalues of a matrix sum

Question

Is it possible to find two matrix $A,B\in L(\mathbb{R}^{n})$, with the eigenvalues of both matrix have negative real part, and the matrix $A+B$ have eigienvalues with non negative real part?

Answer 1

Check out $$ A = \begin{pmatrix}-1&2\\\tfrac 14 &-1\end{pmatrix} \qquad\text{and}\qquad B = \begin{pmatrix}-1&\tfrac 14\\2 &-1\end{pmatrix}. $$ Then both $A$ and $B$ have eigenvalues at $-1\pm\tfrac 1{\sqrt 2} < 0$, but $A+B$ has eigenvalues $-\tfrac{17}4$ and $\tfrac 14$.

Answer 2

If only part of eigenvalues of A, B have negative real part and all eigenvalues of A+B have non-negative real part, $A=\begin{pmatrix}2,0\\0,-1\end{pmatrix},B=\begin{pmatrix}-1,0\\0,2\end{pmatrix}$ is a result. If all eigenvalues of A and B have negative real part, the sum of those eigenvalues of A and B has negative real part or the trace of A+B has negative real part so that at least one eigenvalue of A+B has negative real part.