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

by Segis Izquierdo   Last Updated September 11, 2019 18:20 PM - source

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).

Answers 1

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 $.

Segis Izquierdo
Segis Izquierdo
September 11, 2019 18:00 PM

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