Convexity of $(X, y) \mapsto y^T X^{-1} y$ [closed]

Question

Let $y \in \mathbb{R}^n$, $X \in \mathcal{S}^n_{++}(\mathbb{R})$. Why would function $ f : (X, y) \mapsto y^T X^{-1} y$ be convex?

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I tried with $(X, x) + t.(Y, y)$ with no result. Also, I thought about using the eigenvalues of $X$, without result. Do you have any idea?

Answer 1

This is shown in example 3.4 "Matrix Fractional Function" of Convex Optimization – Boyd and Vandenberghe. The epigraph is transformed via Schur Complement (to handle the matrix inverse) into a convex LMI.

Answer 2

Here is a complete proof.

1. One proves easily that if $S\in{\mathcal S}^n_{++}$ and if $zz^T\prec S$ (in the order between symmetric matrices), then $$\frac12 y^TX^{-1}y\ge z\cdot y-\frac12 {\rm Tr}(SX).$$ Hint: start with the obvious inequality $$\frac12\left( y^TX^{-1}y+z^TXz\right)\ge z\cdot y.$$
2. On the other hand, the equality is achieved by taking $z=X^{-1}y$ and $S=X^{-1}yy^TX^{-1}$.
3. Hence we have $$\frac12 y^TX^{-1}y=\sup\left\{ z\cdot y-\frac12 {\rm Tr}(SX)\right\},$$ where the supremum is taken among those $(z,S)$ such that $zz^T\prec S$.

Now the right-hand side, being the supremum of linear forms of $(y,X)$, is a convex function.