Existence of Solution for $Ax+\alpha g(x)=0$ Let $A$ be an $n \times n$ non-singular matrix and $g:\mathbb{R}^n \to \mathbb{R}^n$. I want to prove that for sufficiently small $\alpha$ the equation $\alpha  g(x) + Ax = 0$ has a solution.
Any suggestion to solve this problem is greatly appreciated.
 A: If $g$ is continuous on some open set $U$ containing the origin, then we can use the
Brouwer fixed point theorem.
Choose some $r>0$ such that $\overline{B}(0,r) \subset U$ and let $M = \sup_{x \in B(0,r)} \|g(x)\| $. Choose $\alpha$ such that $|\alpha| < {r \over \|A^{-1}|| M }$ and let $\phi(x) = -\alpha A^{-1} g(x)$.
Then $\phi:\overline{B}(0,r) \to \overline{B}(0,r)$ and hence has a fixed point.
A: Lauds to copper.hat for producing the result in the most basic of cases, that is, $g \in \mathcal C^0$ is merely assumed continuous.  
In the comments to the main question, our OP Arthur expressed interest in proofs based on the implicit function theorem, so here goes:
I assume that $g(x)$ is differentiable; in fact $g \in \mathcal C^1$ will suffice for the present purposes.
Define
$F: \Bbb R^{n + 1} = \Bbb R \times \Bbb R^n \to \Bbb R^n \tag 2$
by
$F(\alpha, x) = \alpha g(x) + Ax, \; (\alpha, x) \in \Bbb R \times \Bbb R^n; \tag 3$
then
$F(0, 0) = 0(g(0)) + A0 = 0, \tag 4$
and
$D_x F(\alpha, x) = \alpha D_x g(x) + D_x(Ax) = \alpha D_xg(x) + A, \tag 5$
whence
$D_x F(0, 0) = 0(D_xg(0)) + A = A; \tag 6$
since $A$ is non-singular we see that $D_x F(0, 0)$ is invertible as well; therefore the implicit function theorem implies the existence of an interval
$I_\epsilon = (-\epsilon, \epsilon), \tag 7$
and an open ball of radius $\rho > 0$ about $0 \in \Bbb R^n$,
$B(0, \rho) = \{ x \in \Bbb R^n \mid \Vert x \Vert < \rho \} \tag 8$
such that there is a $\mathcal C^1$ function
$x: I_\epsilon \to B(0, \rho), \; x(\alpha) \in B(0, \rho), \tag 9$
such that
$F(\alpha, x(\alpha)) = 0, \alpha \in I_\epsilon; \tag{10}$
using (3), this translates to
$\alpha g(x(\alpha)) + Ax(\alpha) = 0, \; \alpha \in I_\epsilon, \tag{11}$
as was to be shown.  We observe that , in addition to the conclusion required by the question, we have also shown that $x(\alpha) \in \mathcal C^1$, and it further follows from the implicit function theorem that $x(\alpha)$ is unique, and that no other solutions to (3) exist.
