Trouble seeing why this is the dual of an LP $A$ is an $m \times n$ matrix. Using the notation $x=(x_1, \ldots, x_n)$, $z=(z_1, \ldots, z_m)$, and $y=(y_1, \ldots, y_m)$, I'm reading that if the primal LP is
$$
\min 0x_1 + 0x_2 + \cdots + 0x_n + z_1 + z_2 + \cdots +z_m
$$
s.t.
$$
Ax +z\ge \mathbf{1};\\
z_1, z_2, \ldots, z_m \ge 0.
$$
then the dual LP is
$$
\max y_1 + y_2 + \cdots + y_m$$
s.t.
$$
A^Ty = \mathbf{0}\\
0 \le y_1, y_2, \ldots, y_m \le 1
$$
I don't see where the constraint $A^Ty = 0$ in the dual is coming from. Shouldn't it be $A^Ty \le 0$ ? Or are these constraints somehow equivalent?
 A: Seems like you should just derive it from first principles. The Lagrangian is
$$L(x,z,y,w) = \mathbf{1}^T z - y^T ( Ax + z - \mathbf{1} ) - w^T z$$
where $y$ and $w$ are my two Lagrange multipliers. The dual problem is
$$\begin{array}{ll} \text{maximize} & g(y,w)\triangleq \inf_{x,z} L(x,z,y,w) \\
\text{subject to} & y,w \succeq \mathbf{0} \end{array}$$
In order to get a better expression for $g(y,w)$, I differentiate $L$ with respect 
to $x$ and $z$, revealing two equations:
$$- A^T y = \mathbf{0}, \quad \mathbf{1} - y - w = \mathbf{0}$$
If either these conditions are not met, the dual function $g(y,w)=\inf_{x,z} L(x,z,y,w)$ is unbounded below. So these are the implicit dual equality constraints. Making those equations explicit yields a cleaned-up dual problem:
$$\begin{array}{ll} \text{maximize} & y^T \mathbf{1} \\
\text{subject to} & A y = \mathbf{0} \\ & y + w = \mathbf{1} \\ & y,w \succeq \mathbf{0} \end{array}$$
Eliminating $w$ gives us
$$\begin{array}{ll} \text{maximize} & y^T \mathbf{1} \\
\text{subject to} & A y = \mathbf{0} \\ & y \preceq \mathbf{1} \\ & y\succeq \mathbf{0} \end{array}$$
which is exactly what you are reading.
