If $A$ and $B$ have non-negative eigen values then can we conclude that $A+B$ has non-negative eigen values? Let $A$ and $B$ be $n\times n$ matrices whose all eigen values are non-negative real numbers. What can I say about the signs of the eigen values of $A+B$ ? First of all is it possible that the eigen values of $A+B$ may not be real?
Of course if $A$ and $B$ are symmetric and have non-negative eigen values then it is easy to see that $A+B$ will also have non-negative eigen values (this is just equivalent to positive semidefiniteness). So I am only interested in the non-symmetric case.
Thank you.
 A: The eigenvalues can be negative or non-real. 
Take
$$A=\begin{bmatrix}1&a\\b&1\end{bmatrix},B=\begin{bmatrix}1&c\\d&1\end{bmatrix}.$$
Then $A$ and $B$ have positive real eigenvalues as long as $0<ab<1$ and $0<cd<1$, but the eigenvalues of $A+B$ are negative if
$$(a+c)(b+d)>4$$
and non-real if $(a+c)(b+d)<0$. 
For the first case to hold, take $a=d=3$ and $b=c=1/4$. For the second, take
$$(a,b,c,d)=\left(3,\frac14,-\frac14,-3\right).$$
A: Take $A=\begin{pmatrix} 1& 2\\0&1\end{pmatrix}$, $B=\begin{pmatrix} 1& 0\\2&1\end{pmatrix}$. Then both have $1$ as unique eigenvalue but $A+B$ has zero as eigen value.
Take $C=\begin{pmatrix} 1& 1\\0&1\end{pmatrix}$, $D=\begin{pmatrix} 1& 0\\-1&1\end{pmatrix}$. Then both have $1$ as unique eigen value but $C+D$ has no real eigenvalue.
Take $E=\begin{pmatrix} 1& 1\\0&2\end{pmatrix}$, $F=\begin{pmatrix} 2& 0\\-1&1\end{pmatrix}$. Then both have $1,2$ as distinct eigen value but $E+F$ has no real eigenvalue.
Take $G=\begin{pmatrix} 1& 5\\0&2\end{pmatrix}$, $H=\begin{pmatrix} 2& 0\\5&1\end{pmatrix}$. Then both have $1,2$ as distinct eigen value but $G+H$ has $8,-2$ as eigenvalue.
