Matrix Algebra, Signs of solution I have a system $AX = B$, where $A$, $B$ and $X$ are $N \times N$ matrices. I am interested in the properties of the solution $X$.
$B$ has the following property: the diagonal terms are strictly negative and the off-diagonal terms are strictly positive. Let's call this property GSP. I want to know if $X$ also satisfies GSP.
Of course, this depends on $A$. I know quite a bit -- I know that $A$ is symmetric and positive definite. This implies that it is invertible, so $X = A^{-1}B$. If $A$ were diagonal, then this would be trivial, but I know that $A$ is not diagonal. Is the fact that $A$ is symmetric and positive definite sufficient to guarantee that $X$ has the same signs as $B$ and hence that it satisfies GSP?
 A: Consider $A = \begin{bmatrix}2&-1\\-1&2\end{bmatrix}$ and $B = \begin{bmatrix}-1&3\\3&-1\end{bmatrix}$. 
Clearly, $A$ is symmetric and has eigenvalues $1$ and $3$. Thus, $A$ is positive definite. Also, $B$ is GSP. 
However, $X = A^{-1}B = \dfrac{1}{3}\begin{bmatrix}1&5\\5&1\end{bmatrix}$ is not GSP.
A: Of course not.  Examples are very easy to find.
In fact, consider any invertible matrix $A$ that is not diagonal.  Then $A^{-1}$ is not diagonal; say its element $(A^{-1})_{ij} = s \ne 0$. 
If $s > 0$, consider the matrix element $X_{ij} = (A^{-1} B)_{ij}$ as a linear function
of $B_{jj}$, all other elements of $B$ being held fixed: $f(B_{jj}) = c + s B_{jj}$.  If we decrease $B_{jj}$, $B$ still satisfies GSP, and $X_{ij}$ is decreased.  Decrease $B_{jj}$ enough, and $X_{ij} < 0$, so $X$ fails to satisfy GSP.   
On the other hand, if $s < 0$, consider $X_{ik}$ as a function of $B_{jk}$, where $j \ne k$: $g(B_{jk}) = d + s B_{jk}$.  This time we can increase $B_{jk}$ and it will decrease $X_{ik}$.  Increase it enough, and again $X$ will fail to satisfy GSP.
