2

If we have a system of constraints given by,

$$Ax \preceq_K b$$

where $K$ is a second-order cone, would this simply be the same as requiring that:

$$\|Ax\|_2 \leq b$$

where $\|\cdot\|_2$ is the $2$-norm. Or is there something deeper that I'm missing?

EDIT: Having thought about it a bit more, if we define a second-order cone to be $Q^n=\{(u_0,u_1) \in R \times R^{n-1}|u_0 \geq \|u_1\|_2\}$, would the constraint given by, $$Ax \preceq_{Q^n} b$$ imply that we require $$b_0-(Ax)_0 \geq \|b_1-(Ax)_1\|_2$$

InquisitiveInquirer
  • 221

1 Answers1

3

I don't think they're the same. $Ax \preceq_K b \iff Ax - b \preceq_K 0 \iff -(Ax-b) \in K$. So if $K$ were the positive orthant, then we'd have the standard affine constraint $Ax - b \leq 0$.

A second-order (Lorentz) cone is defined by $L_n := \{(x,t): \|x\|_2 \leq t, t\geq 0, x \in \mathbb{R}^n \}$, so $\tilde{x} = (x,t) \in L_n \iff \tilde{x} \succeq_{L_n} 0$, i.e., $\|x\|_2 \leq t$. Basically, you need to specify the "$t$" in $b-Ax \succeq_{L_n} 0$ (or it seems to be somehow implicitly defined as the last element?).

Note that $ \begin{bmatrix} Ax \\ b \end{bmatrix} \succeq_{L_n} 0 \iff \|Ax\|_2 \leq b $ where $b\in \mathbb{R}$. Also note that if we define $y = Ax$, then $\|y\|_2 \leq b$ and $\|-y\|_2 \leq b$, so we can see that for a "proper" cone, we need the $b$ to scale too, otherwise $y$ and $-y$ are both in the cone.

jjjjjj
  • 2,779