Making the entries of a matrix positive. Suppose I have an invertable $n \times n$ matrix, $M$, where certain entries are negative. Does there exist another invertible matrix, $P$, which i can multiply with $M$ to obtain a matrix will all positive entries?
And a followup:
Suppose I have some $n \times n$ matrix, $M$, where certain entries are negative. Does there exist another matrix, $P$, which i can multiply with $M$ to obtain a matrix will all positive entries i.e. $PM$ = abs($M$)? In cases where it does exist, how do i find it?
 A: 
Edit: as noted by Geoff Robinson, my solution does not meet the OP's requirement that the matrix $P$ should be invertible. As given below, $P$ is not necessarily invertible since ${\rm abs}(M)$ need not be invertible, even if $M$ is.

If $M$ is invertible, then matrix $P$ exists. For example,
$$ P = {\rm abs}(M)M^{-1}$$
as I illustrated in my comments. But computationally, constructing $P$ in the trivial way (${\rm abs}(M)M^{-1}$) is funny because you already computed ${\rm abs}(M)$ in order to construct $P$.
If $M$ is non-invertible, then pseudo-inverse $M^{*}$ can help you in certain cases. For example, if the columns of $M$  are orthogonal, then left inverse $M^{*}M= I$ and $P$ can be constructed. Similarly if rows of $M$  are orthogonal, and right inverse is used.
A: If you want all entries of $PM$ to be positive, then all you have to do is to find an invertible matrix $X$ (of the same size) with all entries positive. Then take $P = XM^{-1}$ and we have $PM = X.$ There is such a matrix $X$. For example, Take $X = I +J,$ where every entry of $J$ is $1.$ Since $J$ is symmetric with trace $n$ and rank $1$, its eigenvalues are $n$ (with multiplicity $1$) and $0$ with multiplicity $n-1.$ Hence $X$ has determinant $n+1$, and is invertible with all entries positive.
