Proving that $(x,Q) \mapsto \langle x,Q^{-1}x\rangle$ is convex

I'm studying

$$ F( (x,Q) ) = \langle x,Q^{-1}x\rangle $$

considered over $\mathbb{R}^n\times\mathbb{R}^{n\times n}_+$, where $\mathbb{R}^{n\times n}_+$ is the set of all positive definite $n\times n$ matrices. Here, $\langle x,y\rangle$ stands for the dot product.

I'm trying to prove that $F$ will be convex when considered over this set. That is, it's obvous that $F$ is convex in $x$ with fixed $Q$ (and in $Q$ with fixed $x$), but the trick is to prove that it will be convex with respect to the pair $(x,Q)$. It can be straightforwardly checked for $n=1$ (considering the Hessian), but how can it be done for an arbitrary $n$?

Solution 1:

This seems to be a little trickier than I thought first!

I'll be using the following lemma:

A function $f$ is convex if and only if its epigraph $$\text{epi}(f) = \{(p, t) | \; p \; \text{in domain of} \;f \; \text{and} \; f(p) \leq t\}$$

is a convex set.

Let $f(x, Q) = x^T Q^{-1}x$ with domain $\mathbb{R}^n \times S^n_{++}$ where $S^n_{++}$ is the set of $n$ by $n$ symmetric positive definite matrices over $\mathbb{R}.$ Now, its epigraph is

$$\{((x, Q), t) | \; Q \succ 0, \; x^TQ^{-1}x \leq t\}\stackrel{\text{Schur Complement}}{=}\{((x, Q), t) | \; Q \succ 0, \; \begin{pmatrix} Q & x \\ x^T &t \end{pmatrix} \succeq 0\}.$$ Last condition is an LMI in $(x,Q, t),$ so is convex.