Content deleted Content added
By the definition of convex cones, their intersection can also be shown to be a convex cone, although not necessarily one that can be defined by a single second-order inequality. |
→Second-order cone: indenting |
||
Line 23:
The standard or unit '''second-order cone''' of dimension <math>n+1</math> is defined as
:<math>\mathcal{C}_{n+1}=\left\{ \begin{bmatrix} x \\ t \end{bmatrix} \Bigg| x \in \mathbb{R}^n,
t\in \mathbb{R}, \|x\|_2\leq t \right\}</math>.
Line 30:
The set of points satisfying a second-order cone constraint is the inverse image of the unit second-order cone under an affine mapping:
:<math>\lVert A_i x + b_i \rVert_2 \leq c_i^T x + d_i \Leftrightarrow
\begin{bmatrix} A_i \\ c_i^T \end{bmatrix} x + \begin{bmatrix} b_i \\ d_i \end{bmatrix} \in
\mathcal{C}_{n_i+1}</math>
Line 38:
The second-order cone can be embedded in the cone of the positive semidefinite matrices since
:<math>||x||\leq t \Leftrightarrow \begin{bmatrix} tI & x \\ x^T & t \end{bmatrix} \succcurlyeq 0,</math>
i.e., a second-order cone constraint is equivalent to a linear matrix inequality (Here <math>M\succcurlyeq 0 </math> means <math>M </math> is semidefinite matrix). Similarly, we also have,
:<math>\lVert A_i x + b_i \rVert_2 \leq c_i^T x + d_i \Leftrightarrow
\begin{bmatrix} (c_i^T x+d_i)I & A_i x+b_i \\ (A_i x + b_i)^T & c_i^T x + d_i \end{bmatrix} \succcurlyeq 0</math>.
| |||