# Proof that a continously differentiable, stable map is has an everywhere-invertible derivative matrix

I apologize for the lack of TeX, I'm new to stack exchange. I am trying to solve the problem: Let U be an open subset of R" and suppose that the continuously differentiable mapping F: U ->- R^n is stable. Prove that at each point χ in U, the derivative matrix DF(x) is invertible. (Hint: Use the First-Order Approximation Theorem.)

By stable, this text (Fitzpatrick), means that for x,y in U, there exists c >0 st. ||F(x)- F(y|| > c||x-y||.

The first-order approx. theory gives us that at any point x in U, 0 = lim h->0 ||F(x+h)-F(x) -DF(x)h|| / ||h||. ||F(x+h)-F(x) -DF(x)h|| / ||h|| > (||F(x+h) - F(x)|| - ||DF(x)h||)/ ||h|| by triangle inequality > c - ||DF(x)h||/||h|| by the stability property.

So, lim h-> 0 ||DF(x)h||/||h|| = c >0.

But that's where I get stuck.... how does that last property imply that DF(x) is invertible?

• Nevermind, I solved this myself, after reviewing a little linear algebra. – Sorry for horrible formatting Apr 2 '14 at 5:34