Winding Number of Linear Map

Question

Let $f: \mathbb{R}^2 \to \mathbb{R}^2$ be a linear map, represented by a matrix $A \in \mathbb{R}^{2 \times 2}$, with non-zero determinant.

Further, define the path $\gamma:[0;1]\to \mathbb{R}^2$, $$\gamma(t):=f\Big(\big(\cos{(2\pi t)},\sin{(2\pi t)}\big)\Big)$$

I must find the winding number of $\gamma$ at $0$.

We have seen that $$ W(\gamma,0)=\frac{1}{2\pi}\int_\gamma \frac{-ydx+xdy}{x^2+y^2} $$ but I'm not sure about how to apply this formula.

Thanks in advance!

Answer 1

It is doubtful however that $\gamma$ takes values in $\mathbb{S}^1$ if $A$ is an arbitrary matrix. So, there is some information that you probably forgot to mention.

Anyway...

If $A=\begin{pmatrix}a & b \cr c & d\end{pmatrix}$, then $f((x,y))=(ax+by,cx+dy)$.

Next, write $\gamma(t)=(x(t),y(t))$ in terms of $a,b,c,d$, and just use the definition of an integral along a path $\gamma$. You will be reduced to compute an standard integral of a function in the variable $t$.