Positive vector times stable matrix If all eigenvalues of $A$ have negative real part, and $w,v$ are positive vectors (positive entries), does the following inequality hold?
$$w^TAv < 0$$ 
 A: Counterexample:
$$\left(\begin{matrix}
4& 1
\end{matrix}\right)
\left(\begin{matrix}
-1 & 1\\0 & -2
\end{matrix}\right)
\left(\begin{matrix}
1\\4
\end{matrix}\right)=4$$
Where the matrix is
$$\left(\begin{matrix}
-1 & 1\\0 & -2
\end{matrix}\right)
=
\left(\begin{matrix}
-1 & -1\\0 & 1
\end{matrix}\right)
\left(\begin{matrix}
-1 & 0\\0 & -2
\end{matrix}\right)
\left(\begin{matrix}
-1 & -1\\0 & 1
\end{matrix}\right)^{-1}
$$

By using vectors $e_i=(0,\dots,0,1,0,\dots,0)^T$ (where the only nonzero entry is the $i$-th one), you can prove that all entries of your matrix $A$ must be nonpositive. That's because $e_iAe_j^T=a_{ij}$. To have positive vectors, replace the zeros by $\varepsilon>0$ and let $\varepsilon\to0$, then you must have $a_{ij}\le0$. This condition is necessary, but not sufficient, since for instance the null matrix does not satisfy your condition either. But if at least one entry $a_{ij}$ of $A$ is negative, this works because we know that $u$ and $w$ must not have zero entries, hence the corresponding $\lambda e_iA\mu e_j^T<0$, and all the other terms in $uAw^T$ are $\le0$.
But note that this can happen even if $A$ has positive eigenvalues. For instance the eigenvalues of $\left(\begin{matrix}
-1 & -2\\-2 & -1
\end{matrix}\right)
$ are $-3$ and $1$.
