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:<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]].
:<math>x^T M x \geq 0 \text{ for all } x \in \mathbb{R}^n</math>
which is not a linear inequality in the conventional sense.
Similarly, we also have,
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