I am trying the following exercise :

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

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

Have you an idea  ?

Thank you,

Orso

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?

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?

I am trying the following exercise :

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

I tried with $(X, x) + t.(Y, y)$ with no result. Also, i thought about using the eigenvalues of $X$ without result...

Have you an idea ?

Thank you,

Orso

I am trying the following exercise :

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

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

Have you an idea ?

Thank you,

Orso