The positive rank of a square matrix is defined in Theorem $3$ of Expressing Combinatorial Optimization Problems by Linear Programs by Mihalis Yannakakis as follows: given a $n\times n$ matrix $A$, the positive rank $rank_{\Bbb R}^+(A)$ is the smallest $m$ such that $A=LR$ for a non-negative $n\times m$ matrix $L$, and non-negative $m\times n$ matrix $R$.This concept is valuable in communication complexity, since it was shown that if $rank_{\Bbb R}^+(A)$ and $rank_{\Bbb R}(A)$ could be subexponentially related for a $0/1$ matrix $A$, then the log-rank conjecture holds.Is there an exponential separation between $rank_{\Bbb R}^+(A)$ and $rank_{\Bbb R}(A)$ for a general non-negative real matrix $A$ (as opposed to just $0/1$) or is this problem also open?I checked the references in Jukna's book, but I am still unable to clarify the above question.
On the notion of positive rank of a matrix
communication complexity;definitions
There are examples of $n\times n$ real matrices of rank at most $3$ and non-negative rank at least $\sqrt{2n}$. So the non-negative rank cannot be bounded by any function of the rank in general. The construction I am aware of goes through extension complexity. An explanation follows.The extension complexity $xc(K)$ of a convex set $K$ in $\mathbb{R}^d$ is the smallest possible number of facets of a (higher-dimensional) polytope that projects to $K$. (Equivalently the smallest possible number of constraints in a linear program for $K$). A famous theorem of Yannakakis shows that if there exist $A$, $b$ and $v_1, \ldots, v_n$ such that $K = \{x: Ax \leq b\} = \mathrm{conv}(v_1, \ldots, v_n)$, then $$xc(K) = \mathrm{rank}^+(S),$$where $S$ is the slack matrix, defined by $S = b1^T - A^TV$ for $V = (v_1, \ldots, v_n)$, i.e. the matrix whose columns are the $v_i$.A result by Fiorini, Rothvoss, and Tiwari shows that there exist two-dimensional $n$-gons with extension complexity at least $\sqrt{2n}$. Since any $n$-gon $P$ in $\mathbb{R}^2$ can be written as $P = \{x: Ax \leq b\} = \mathrm{conv}(v_1, \ldots, v_n)$ for $A$ an $n\times 2$ matrix, $b$ an $n\times 1$ vector and $V = (v_1, \ldots, v_n)$ a $2\times n$ matrix, the slack  matrix $S = b1^T - A^TV$ of $P$ is the sum of a rank $1$ and a rank $2$ matrix, and therefore is of rank at most $3$. On the other hand by Yannakakis's and Fiorini et al.'s results $\mathrm{rank}^+(S) = xc(P) \geq \sqrt{2n}$. 