I have the following porblem.
Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be a linear mapping, represented by a $2\times 2$ matrix. Moreover let the determinant be nonzero. Show that the local degree of $f$ at the origin is $+1$ if the determinant is positive and $-1$ if it is negative.
I wanted to start as follows.
Let $A$ be te matrix that represents the linear map $f$ s.t. $\det(A)\neq 0$ and $f(x,y)=A\cdot (x,y)$. Let $U\in \mathfrak{U}(0,0)$ be a neightburhood of $(0,0)$ then it is clear that $f(0,0)\neq f(x,y)$ for all $(x,y)\in U\setminus \{(0,0)\}$ since $f$ is bijective. Now we know that $f$ restricts to a continuous function $F:\mathbb{S}^1\rightarrow \mathbb{R}^2\setminus \{(0,0)\}$. Then by def $$deg_{(0,0)}(f)=W(F|_{\mathbb{S}^1},(0,0))$$Where $W$ is the winding number
But I somehow from this point I don't know how to procede. Could someone help me how we can rewrite $F$ maybe?
Thanks for your help.