# Positive-Eigenvalue Jacobian $\Rightarrow$ Invertible?

Suppose $$f : \mathbb{R}^n \to \mathbb{R}^n$$ has Jacobian $$Jf : \mathbb{R}^n\to\mathsf{M}_n(\mathbb{R})$$ with positive eigenvalues everywhere. Is $$f$$ (globally) injective (invertible on its range)?

If $$Jf$$ was also guaranteed to be symmetric, this would be true by this question. We also know $$f$$ is locally invertible by the inverse function theorem.

• Take $f=\arctan$. Then $f' >0$ but it is not surjective. Commented Aug 7, 2018 at 0:34
• @copper.hat I guess I should specify invertible on it's range i.e. injective Commented Aug 7, 2018 at 0:35
• There was nothing about the linked proof which required symmetry. All you need is that $\langle (Jf)x,x\rangle >0$ for all $x\neq 0$, which follows since the smallest eigenvalue is $\min_{\|x\|=1} \langle (Jf)x,x\rangle$. Commented Aug 7, 2018 at 0:59
• @MikeEarnest Forgive me if I'm missing something, but I'm pretty sure that only works for self-adjoint $Jf$. $A = \begin{bmatrix}1&-3\\0&1\end{bmatrix}$ is a counterexamples with $x=[1,1]^*$. Commented Aug 7, 2018 at 1:03
• You are right, apologies for my hasty comment. Commented Aug 7, 2018 at 1:09