Given a rank $2r$ matrix $M\in\Bbb Q^{n\times n}$ can we find two matrices $M_+\in\Bbb Q_{\geq0}^{n\times n}$ and $M_-\in\Bbb Q_{\geq0}^{n\times n}$ each of rank $r$ such that $M=M_+-M_-$ holds?

Though $M=AB$ where $A,B'\in\Bbb Q^{n\times r}$ are rank $r$ and we can rewrite this as $$M=(A_+B'_++A_-B'_-)-(A_+B'_-+A_-B'_+)$$ where $A_+,A_-,B'_+,B'_-\in\Bbb Q_{\geq0}^{n\times r}$ holds it is unclear $$rank(A_+B'_++A_-B'_-)=rank(A_+B'_-+A_-B'_+)=r$$ holds.

Given a rank $2r$ matrix $M\in\Bbb Q^{n\times n}$ can we find two matrices $M_+\in\Bbb Q_{\geq0}^{n\times n}$ and $M_-\in\Bbb Q_{\geq0}^{n\times n}$ each of rank $r$ such that $M=M_+-M_-$ holds?

Though $M=AB$ where $A,B'\in\Bbb Q^{n\times 2r}$ are rank $2r$ and we can rewrite this as $$M=(A_+B'_++A_-B'_-)-(A_+B'_-+A_-B'_+)$$ where $A_+,A_-,B'_+,B'_-\in\Bbb Q_{\geq0}^{n\times 2r}$ holds it is unclear even $$rank(A_+B'_++A_-B'_-)=rank(A_+B'_-+A_-B'_+)=2r$$ holds while I seek $M_+-M_-$ as rank $r$ which is minimum possible.

Given a rank $2r$ matrix $M\in\Bbb Q^{n\times n}$ can we find two matrices $M_+\in\Bbb Q_{\geq0}^{n\times n}$ and $M_-\in\Bbb Q_{\geq0}^{n\times n}$ each of rank $r$ such that $M=M_+-M_-$ holds?

Though $M=AB$ where $A,B'\in\Bbb Q^{n\times r}$ are rank $r$ and we can rewrite this as $$M=(A_+B_++A_-B_-)-(A_+B_-+A_-B_+)$$ where $A_+,A_-,B_+,B_-\in\Bbb Q_{\geq0}^{n\times r}$ holds it is unclear $$rank(A_+B_++A_-B_-)=rank(A_+B_-+A_-B_+)=r$$ holds.

Given a rank $2r$ matrix $M\in\Bbb Q^{n\times n}$ can we find two matrices $M_+\in\Bbb Q_{\geq0}^{n\times n}$ and $M_-\in\Bbb Q_{\geq0}^{n\times n}$ each of rank $r$ such that $M=M_+-M_-$ holds?

Though $M=AB$ where $A,B'\in\Bbb Q^{n\times r}$ are rank $r$ and we can rewrite this as $$M=(A_+B'_++A_-B'_-)-(A_+B'_-+A_-B'_+)$$ where $A_+,A_-,B'_+,B'_-\in\Bbb Q_{\geq0}^{n\times r}$ holds it is unclear $$rank(A_+B'_++A_-B'_-)=rank(A_+B'_-+A_-B'_+)=r$$ holds.

Given a rank $2r$ matrix $M\in\Bbb Q^{n\times n}$ can we find two matrices $M_+\in\Bbb Q_{\geq0}^{n\times n}$ and $M_-\in\Bbb Q_{\geq0}^{n\times n}$ each of rank at most $r$ such that $M=M_+-M_-$ holds?

Given a rank $2r$ matrix $M\in\Bbb Q^{n\times n}$ can we find two matrices $M_+\in\Bbb Q_{\geq0}^{n\times n}$ and $M_-\in\Bbb Q_{\geq0}^{n\times n}$ each of rank $r$ such that $M=M_+-M_-$ holds?

Though $M=AB$ where $A,B'\in\Bbb Q^{n\times r}$ are rank $r$ and we can rewrite this as $$M=(A_+B_++A_-B_-)-(A_+B_-+A_-B_+)$$ where $A_+,A_-,B_+,B_-\in\Bbb Q_{\geq0}^{n\times r}$ holds it is unclear $$rank(A_+B_++A_-B_-)=rank(A_+B_-+A_-B_+)=r$$ holds.