Construct example - $A$ and $B$ have strictly positive eigenvalues, but $A+B$ and $AB$ have strictly negative eigenvalues. As a follow-up to this question I asked, I wondered what would happen if I imposed the weaker condition of having positive eigenvalues, rather than being positive definite.
How do I construct an example of two matrices $A$ and $B$ such that:
1) $A$ and $B$ have strictly positive eigenvalues.
2) $A + B$ has strictly negative eigenvalues (is this even possible?).
3) $AB$ has strictly negative eigenvalues.
Generally, I'm unsure how to begin going about constructing an example of a matrix that satisfies these properties.
 A: The trace of a matrix is the sum of its eigenvalues.  Also, $$\mathrm{tr}(A+B) = \mathrm{tr}(A)+\mathrm{tr}(B).$$  If $A$ and $B$ have strictly positive eigenvalues, then 
$$\mathrm{tr}(A),\mathrm{tr}(B)>0\implies \mathrm{tr}(A+B)>0,$$ 
so it is not the case that $A+B$ has negative eigenvalues.
For a case where $A,B$ have only positive eigenvalues and $AB$ has only negative eigenvalues, take $$A=\pmatrix{1&0\\0&4}, B=\pmatrix{4&-3\\3&-2}$$
The eigenvalues of $A$ are obviously $1$ and $4$, both eigenvalues of $B$ are $1$, and both eigenvalues of $AB$ are $-2.$  I constructed this just by messing around with the characteristic polynomials.  I don't have any insight into the problem at all.  
A: if $A$ and $B$  are symmetric and  strictly positive definit, you can not have neither $A+B$ nor $AB$ to be negative definite
A: For an example where $A$ and $B$ have all positive eigenvalues while $AB$ has all negative eigenvalues, consider 
$$ A = \pmatrix{10 & 8\cr -9 & -7\cr},\ B = \pmatrix{1 & 0\cr 0 & 2\cr} $$
$A$ and $B$ both have eigenvalues $1$ and $2$, while $AB$ has a double eigenvalue $-2$.
