Set $\mathcal{NC}:=\{ A \in M_2(\mathbb{R}) \, | \det A \ge 0 \, \,\text{ and } \, A \text{ is not conformal} \,\}$, where by a non-conformal matrix, I refer to a matrix whose singular values are distinct. (i.e. I allow non-zero singular matrices in $\mathcal{NC}$).
Let $A:\mathbb{S}^1 \to \mathcal{NC}$ be a matrix field defined as follows:
Described w.r.t the frame $\{\frac{\partial}{\partial \theta},\frac{\partial}{\partial r}\}$,
$$ A(\frac{\partial}{\partial \theta})=\frac{\partial}{\partial \theta}, A(\frac{\partial}{\partial r})=\frac{\partial}{\partial r}+c\frac{\partial}{\partial \theta},$$
where $c>0$ is a fixed parameter.
Question: Is $A:\mathbb{S}^1 \to \mathcal{NC}$ homotopic to a constant map?
Note that I define the range to be $\mathcal{NC}$. Of course, $A$ is null-homotopic when considering the range to be all $M_2(\mathbb{R})$.
In matrix form, $$[A]_{ \frac{\partial}{\partial r},\frac{\partial}{\partial \theta}}=\begin{pmatrix} 1 & 0 \\\ c & 1\end{pmatrix}.$$
A computation shows that $A$ has the following form w.r.t the standard basis $e_1,e_2$:
$$A(\theta)=[A]_{e_1,e_2}=\begin{pmatrix} 1-\frac{c}{2}\sin(2\theta) & -c\sin^2(\theta) \\\ c\cos^2(\theta) & 1+\frac{c}{2}\sin(2\theta)\end{pmatrix}$$
i.e. this is $A(\theta) \in \mathcal{NC} \subseteq \text{Hom}(\mathbb R^2,\mathbb R^2)$.
I think that $A(\theta)$ may not be contractible, since from $\cos^2(\theta)$ we can compute $\cos(2\theta)$, which together with $\sin(2\theta)$ determines $2\theta$. (and the maps $\mathbb{S}^1 \to \mathbb{S}^1$ given by $\theta \to 2\theta$ is not homotopic to a point. However, I ask about $A$ as a map $\mathbb{S}^1 \to \mathcal{NC}$; in order to get a map $\mathbb{S}^1 \to \mathbb{S}^1$, we would need to compose it with some continuous map $\mathcal{NC} \to \mathbb{S}^1$.
Here is what I know about the topology of $\mathcal{NC}$:
Let $\mathcal D=\{ (\sigma_1,\sigma_2) \, | 0 \le \sigma_1 < \sigma_2\}$. Then the map
\begin{align*} \mu: SO_2\times \mathcal D\times SO_2\to \mathcal{NC}\\ (U,\Sigma,V)\mapsto U\Sigma V^T \end{align*}
is a $2$-fold smooth covering map*. (i.e. $\mu(U,\Sigma,V)=\mu(-U,\Sigma,-V)$, and this is the only ambiguity in $U,V$ for a pre-image of a given point in $\mathcal{NC}$.
Since $SO_2 \cong \mathbb{S}^1$, and since after identifying antipodal points in $\mathbb{S}^1 \times \mathbb{S}^1$, we get the $2$-torus $\mathbb{T}^2$ again, it follows that $\mathcal{NC} \cong \mathbb{T}^2 \times \mathcal D$.
*I am not entirely sure regarding the behaviour at the boundary points where $\sigma_1=0$, but I don't think this creates a serious problem.
