Suppose we have a non singular bilinear application $\mu$:$\mathbb{R}^n \times \mathbb{R}^n \rightarrow\mathbb{R}^n$ with $\mu(v,w)=\mu (w,v)$. Let the continous application $g: S^{n-1} \rightarrow S^{n-1}$ given by $g(x)= \frac{\mu(x,x)}{|| \mu(x,x) || }$.
I need to show that $g(x)=g(y)$ implies $y= \pm x$.
The case where $y$ and $x$ are linearly dependent is easy. Then, we have $y=ax$, thus $1=||y||= |a| ||x||=|a|$. So we have $a= \pm 1$ (Because x,y are in $S^{n-1}$.)
Here is my question: What happens when $x$ and $y$ are linearly independent? Can I directly conclude that we cannot have $g(x)=g(y)$?