How is this a linear matrix inequality?

Question

In example 3.4 of Stephen Boyd & Lieven Vandenberghe's Convex Optimization, it is mentioned that the last condition of

$$\text{epi} = \left\{ (x,Y,t) \mid Y \succ 0, x^T Y^{-1} x \leq t \right\}$$

is a linear matrix inequality (LMI) in $(x,Y,t)$. However the linear matrix inequality is written as (in Eq. 2.11 of same book)

$$A(x) = x_1 A_1 + x_2 A_2 + \cdots + x_n A_n \preceq B$$

where $A_i$ and $B$ are symmetric matrices. How to show that $x^TY^{-1}x\leq t$ is a linear inequality in $(x,Y,t)$? Any help in this regard will be much appreciated.

Answer 1

Using the Schur complement, if ${\bf Y} \succ {\bf O}$ then $t - {\bf x}^\top {\bf Y}^{-1} {\bf x} \geq 0$ is equivalent$^\color{magenta}{\star}$ to the following linear matrix inequality (LMI)

$$ \begin{bmatrix} {\bf Y} & {\bf x} \\ {\bf x}^\top & t\end{bmatrix} \succeq {\bf O} $$

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_{$\color{magenta}{\star}$ Jean Gallier, The Schur complement and symmetric positive semidefinite (and definite) matrices, December 10, 2010.}