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A square matrix $M$ is called a $P$-matrix ($P$ stands for "positive") if all the principal minors of $M$ are positive. My question is the following.

Let $\mathscr{M}$ be a class of $n\times n$ real matrices such that $\forall I \subset \{1,\cdots,n\}, \forall M\in\mathscr{M}$, the principal minor of $M$ corresponding to $I$ has the prescribed sign (depending on $I$). Under what assumption can one guarantee there exist real matrices $A$ and $B$ such that, for all $M\in \mathscr{M}$, the matrix $AMB$ is a $P$-matrix?

The assumptions I was expecting are upper bound on the entries and lower bound on the principal minors. But there is no reason to exclude other possibilities.

Thanks!

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The principle behind my answer still stands for the modified question: for $M$ of less then full rank (i.e., not invertible) the same will be the case for $AMB$, which therefore cannot be a $P$ matrix; on the other hand for $M$ invertible one can make $AMB$ the identity which is certainly a $P$ matrix. The fact that all invertible matrices form just a single class for the relation of being equivalent matrices makes such questions about positivity uninteresting. –  Marc van Leeuwen Feb 15 '12 at 22:28
    
Well I guess you miss an important point in my question.. The matrices $A$ and $B$ are required make $M$ into a P-matrix for EVERY $M\in \mathscr{M}$. –  Syang Chen Feb 16 '12 at 0:28
    
That interpretation was not clear to me: saying "there exist $x$ such that $prop(x,y)$, for all $y$" could mean either "for all $y$ there exist $x$ such that $prop(x,y)$", or "there exist $x$ such that for all $y$ one has $prop(x,y)$". I've edited the question such that the latter (apparently intended) interpretation is clear. It does not make things a lot more interesting though: since now $A$ and $B$ must be fixed throughout, one can take $\mathscr M=\{ A^{-1}XB^{-1}\mid X\in P' \}$ where $P'$ is any subset of the $P$-matrices, possibly all of them. –  Marc van Leeuwen Feb 16 '12 at 8:14
    
Thanks for editing. –  Syang Chen Feb 16 '12 at 22:58
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1 Answer 1

Since the determinant of $M$ is one of its principal minors, it is non-zero, and $M$ is invertible. Take $A$ to be its inverse and $B$ the identity, then $AMB$ is also the identity, and it is a $P$-matrix. Now what was the question you really wanted to ask?

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Thank you for pointing that out. Please see the edit. –  Syang Chen Feb 15 '12 at 19:07
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