How to represent polyhedron as the inverse image of the Cartesian product of the nonnegative orthant and the origin?

37 Views Asked by At

In Convex Optimization by Boyd and Vandenberghe, I don't understand how the formulation for Example 2.9 works.

Example 2.9: Polyhedron. The polyhedron $ \{ x | Ax \preceq b, Cx=d \}$ can be expressed as the inverse image of the Cartesian product of the nonnegative orthant and the origin under the affine function $f(x) = (b-Ax, d-Cx)$ $$ \{ x | Ax \preceq b, Cx=d \} = \{ x | f(x) \in \mathbb{R}^m_+ \times \{0\} \} $$

If I understand correctly, $\mathbb{R}^m \times \{ 0 \}$ looks something like $(x,y,0)$ if $m=2$.

I'm assuming that $Cx=d$ is reformatted to $d-Cx=0$, which can represented matrix-wise as $[d-Cx | 0]$, which kind of makes sense to me.

How then, does $b-Ax \succeq 0$ work? Since $b-Ax$ produces a vector, would this be vector based (component-wise inequality) comparison and not positive semidefinite? How do you represent that in the form of $f(x)=(z_1,...,z_m,0)$?

And how does the original equation represent both inequalities at the same time? I suspect that I'm missing some background/definitions with notation.