We were given a question in my multivariable calculus that I can't seem to get to the bottom of.
The question goes:
Let ${F}: \mathbb{R}^n \to \mathbb{R}^n$ be such that $F(X)$ is a unit vector for all $ x\in\mathbb{R}^n$. Prove that $\det( F'(x)) = 0$ for all $ x \in \mathbb{R}^n$.
I've gotten as far as showing that $F^T.F' = 0$ but I'm not sure where to go from here. How do I relate the determinant to this? I strongly suspect the shortfall in understanding is coming from linear algebra. I have also tried taking the transpose of the two sides but I'm not sure how to relate this to the determinant of the derivative function.
(Apologies for not using the proper math format on the site; please note that $F$ is a vector-valued function mapping from $\mathbb{R}^n \to \mathbb{R}^n$)
If $F^T(x) F'(x) = 0$ then $F'(x)$ must be singular, hence $\det F'(x) = 0$.