The problem is:
$(y, t) \in \mathcal{L}_{2}^{n+1} \iff \quad
\begin{bmatrix}
t & y \\
y^{t} & tI_{n}
\end{bmatrix} \in \mathcal{S}_{+}^{n+1}$
With $\mathcal{L}_{2}^{n+1} = \{(x, t) \in \mathbb{R}^{n+1} |\left\lVert x\right\rVert _{2} \leq t \}$ (the $\textit{Lorentz Cone}$ or the $\textit{Second-order Cone}$ )
where $\left\lVert x\right\rVert_{2}\leq t \iff
\quad
\begin{pmatrix}
y , t \\
\end{pmatrix}\quad
\begin{bmatrix}
I & 0 \\
0 & -1
\end{bmatrix} \quad
\begin{pmatrix}
y \\
t
\end{pmatrix} \leq 0$.
and $\mathcal{S}_{+}^{n+1} = \{A\in \mathbb{R}^{n+1}| x^{T}Ax \geq 0\}$ is the set of all the semi-definite matrix with dimension $(n+1)\times(n+1)$.
I think that we can use the matricial caracterization for the elements in the Lorentz Cone but i honestly i don't know how it follows. Thanks.
Best Answer
Suppose $\|x\| \le t$. Then consider \begin{eqnarray} \begin{bmatrix} a & b^T \end{bmatrix} \begin{bmatrix} t & x^T \\ x & tI\end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} &=& a^2 t + 2 a x^T b + t \|b\|^2 \\ &\ge & a^2 t - 2 a \|x\|\|b\| + t \|b\|^2 \\ &\ge & a^2 \|x\| - 2 a \|x\|\|b\| + \|x\| \|b\|^2 \\ &=& \|x\| (a-\|b\|)^2 \end{eqnarray} from which positive semi definiteness follows.
In the other direction, consider: \begin{eqnarray} \begin{bmatrix} \|x\| & -x^T \end{bmatrix} \begin{bmatrix} t & x^T \\ x & tI\end{bmatrix} \begin{bmatrix} \|x\| \\ -x \end{bmatrix} &=& 2\|x\|^2(t-\|x\|) \end{eqnarray} from which $t \ge \|x\|$ follows.