# Is strict complementary slackness unique?

We have $$A \in \mathbb{R}^{m \times n}$$ and rank$$(A)=m$$.

By strict complementary slackness, we know that if the primal $$(P)$$ (in standard equality form) has an optimal solution, then the dual $$(D)$$ also has an optimal solution. Furthermore, there exists optimal solutions $$\bar{x}$$ and $$\bar{y}$$ such that for every index $$j \in \{1, 2, \dots, n\}$$, either $$\bar{x}_j = 0$$ or $$(A^T\bar{y}-c)_j = 0$$, but not both.

My question is, do these $$\bar{x}$$ and $$\bar{y}$$ uniquely define a partition of the columns of $$A$$ (i.e. the zero elements)? Or is it possible to have, say, two such sets $$\bar{x_1}, \bar{y_1}$$ and $$\bar{x_2}, \bar{y_2}$$ such that $$\bar{x_1}$$ and $$\bar{x_2}$$ have different zero elements?