Convexity of $(x, X) \mapsto x^\top X^{-1} x$

Asked Sep 14 '21 at 20:56

Active Sep 15 '21 at 06:33

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Is the function $f: \mathbb{R}_+^n \times \mathcal{S}_{++}^n$ defined by $$f(x, X) = x^\top X^{-1} x$$ convex on $\mathbb{R}_+^n \times \mathcal{S}_{++}^n$?

Note: $\mathbb{R}_+^n$ is the set of component wise non-negative vectors (the non-negative orthant), and $\mathcal{S}_{++}^n$ is the set of positive definite $n\times n$ matrices.

What I’ve tried

$f$ is an inner product induced by $X^{-1}$, and $X^{-1}$ is a positive definite matrix too. $f$ is convex on $\mathbb{R}_+^n$ because it is convex on all of $\mathbb{R}^n$, keeping $X$ fixed. However if $X$ is also allowed to vary then that proof doesn’t apply.

I’m thinking along the lines of: How do we characterise / compute a term like $(\lambda X + (1 - \lambda) Y)^{-1}$?

edited Sep 15 '21 at 06:33

Rodrigo de Azevedo

asked Sep 14 '21 at 20:56

Sean

Take a look at this. – Rodrigo de Azevedo Sep 15 '21 at 05:46
This is not a quadratic form. – Rodrigo de Azevedo Sep 15 '21 at 06:33
1

@RodrigodeAzevedo The duplicate has misleading title, it is the same question as here. – daw Sep 15 '21 at 12:20
@daw It is not an exact duplicate, though, as here $x$ is in the non-negative orthant. – Rodrigo de Azevedo Sep 15 '21 at 12:28
@daw that was what I needed thank you! What I don’t understand is why the solution set of $\begin{pmatrix} X & x \ x^\top & t \end{pmatrix} \succeq 0$ would be a convex set. – Sean Sep 15 '21 at 18:58
@Sean Because it is a spectrahedron. – Rodrigo de Azevedo Sep 15 '21 at 19:04

Convexity of $(x, X) \mapsto x^\top X^{-1} x$

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