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The function $f: R^n \times S^n \rightarrow R $ defined as $ f(x,y)=x^TY^{-1}x$ is convex on $\operatorname{dom} f = R^n \times S_{++}^n$. One easy way to establish convexity of f is via its epigraph:

$$ \operatorname{epi} f = \{(x,Y,t)\mid Y \succ 0, x^TY^{-1}x \leq t \} = \{ (x,Y,t) \mid \begin{bmatrix} Y & x \\ x^T & t \end{bmatrix} \succeq0, Y\succ 0 \} $$

using the Schur complement condition for positive semidefiniteness of a block matrix.The last condition is a linear matrix inequality in $(x, Y, t)$, and therefore epi f is convex.

This example is in my book but i don't understand it. I know what epigraph is and I know why $Y \succ 0, x^TY^{-1}x \leq t$ should hold. But I don't understand the matrix. How the matrix semidefiniteness would help us? And how to obtain such matrix?

Michael Hardy
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lino
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  • What is $S^n$? Surely not the $n$-sphere, as it does not contain matrices ... and what is $S_{++}^n$? – martini Oct 02 '15 at 08:07

1 Answers1

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As usual, let $\mathbb{R}^{m \times n}$ be the set of all $m$-by-$n$ real matrices and $S^n_{++} \subset \mathbb{R}^{n \times n}$ denote the set of positive-definite $n$-by-$n$ real matrices. Now, using Proposition 2.1(2) of this paper with $C:=Y \in S^n_{++}$, and $B := x^T \in \mathbb{R}^{1 \times n}$, and $A = t \in \mathbb{R}^{1 \times 1}$, we have \begin{eqnarray} \begin{split} \textrm{epi}f &= \left\{(x,Y,t) \in \mathbb{R}^n \times S^n_{++} \times \mathbb{R}| t-x^TY^{-1}x \geq 0\right\}\\ &= \left\{ (x,Y,t) \in \mathbb{R}^n \times S^n_{++} \times \mathbb{R}| \begin{bmatrix} t & x^T \\ x & Y \end{bmatrix} \succeq 0 \right\}, \end{split} \end{eqnarray} which is a convex set (linear matrix inequalities).

dohmatob
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