# Invertibility of function $x \circ (Ax)$ ($\circ$ for Hadamard product)

I would like help with the conjecture that the function $$f:\mathbb{R}_{\ge 0}^n\to\mathbb{R}_{\ge 0}^n$$ with $$f(x) = x \circ (Ax)$$ where

∘ is the Hadamard product (equivalently, $$f_i(x) = x_i \sum_j a_{ij}x_j$$ )

and where A is a symmetric real matrix with elements $$a_{ij} \ge 1$$,

is invertible (locally would be enough, though I suspect globally on its domain, nonnegative components).

I did not get anywhere looking at the Jacobian or trying to show that f is one-to-one.

The function is such that for a non negative scalar k, $$f(kx) = k^2 f(x)$$, so it maps lines starting at the origin to lines starting at the origin (more precisely, rays in the nonnegative orthant with endpoint at the origin).

• The last observation $f(kx)=k^2f(x)$ implies $f(\pm kx)=k^2 f(x)$, so $f$ is not injective and cannot be globally invertible. – lisyarus Sep 4 at 17:22
• Thanks, I am looking for invertibility in its domain, non negative components. – Segis Izquierdo Sep 4 at 21:20
• Oh, I didn't notice the $\geq 0$ part in $\mathbb R^n_{\geq 0}$. I apologize. – lisyarus Sep 4 at 22:50
• You may also want $A$ to be invertible, otherwise there is some $x \neq 0$ with $Ax=0$ and thus $f(x)=0$, contradicting global invertibility (yet it may happen that $x \not \in \mathbb R^n_{\geq 0}$, so something less restrictive than invertibility may work). Maybe a good starting point would be using the spectral decomposition for $A$: symmetric matrices always have an orthogonal eigenbasis with real eigenvalues. I'm not sure whether an orthonormal basis would work nicely with Hadamard product, though. – lisyarus Sep 4 at 22:54
• Thank you. For the first part, as the elements of $A$ are positive and the components of $x$ are non negative, the only $x$ such that $Ax = 0$ would be $x = 0$. I will be thinking about your second lead. – Segis Izquierdo Sep 5 at 6:47

For $$x_i > 0$$ it can be shown that the Jacobian is strictly diagonally dominant, and consequently non-singular. With some care, local invertibility can then be proved in $$\mathbb{R}_{\ge 0}^n$$.