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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.

Rodrigo de Azevedo
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Frank Moses
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1 Answers1

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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)

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

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

Rodrigo de Azevedo
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  • Thank you for your answer. But in the book the definition of LMI is as given in the second equation of my post. – Frank Moses May 31 '18 at 09:40
  • I mean how the generalized inequality that you mentioned in your answer turns into the second generalized inequality in my post – Frank Moses May 31 '18 at 09:43
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    If $\rm Y$ is $2 \times 2$, then $$\begin{align} \begin{bmatrix} y_{11} & y_{12} & x_1\ y_{12} & y_{22} & x_2\ x_1 & x_2 & t\end{bmatrix} &= y_{11} \begin{bmatrix} 1 & 0 & 0\ 0 & 0 & 0\ 0 & 0 & 0\end{bmatrix} + y_{12} \begin{bmatrix} 0 & 1 & 0\ 1 & 0 & 0\ 0 & 0 & 0\end{bmatrix} + y_{22} \begin{bmatrix} 0 & 0 & 0\ 0 & 1 & 0\ 0 & 0 & 0\end{bmatrix} +\\\ &,, + x_{1} \begin{bmatrix} 0 & 0 & 1\ 0 & 0 & 0\ 1 & 0 & 0\end{bmatrix} + x_{2} \begin{bmatrix} 0 & 0 & 0\ 0 & 0 & 1\ 0 & 1 & 0\end{bmatrix} + t \begin{bmatrix} 0 & 0 & 0\ 0 & 0 & 0\ 0 & 0 & 1\end{bmatrix}\end{align}$$ – Rodrigo de Azevedo May 31 '18 at 09:48