Let $\langle \cdot | \cdot\rangle$ be the standard inner product over $\mathbb{R}^{2}$ and let $A \in M_{2}(\mathbb{R})$ be a matrix such that $\|Av\| = \|v\|$ for every $v \in \mathbb{R^{2}}$ and such that $\det(A) < 0$. Show that $A$ is a reflection matrix.
I started doing it by showing that $\det(A) = \pm1$. If $\det(A) = 1$, $A$ is a rotation matrix. However, in our case, it is given that $\det(A)<0$ and, therefore, it must be $\det(A) = -1$.
I also showed that there are exists $x, y \in \mathbb{R}$ such that $x^{2} + y^{2} = 1$ and such that
$$A = \begin{bmatrix} x & y \\ y & -x \end{bmatrix}$$
Now I'm stuck on this: find $a, b \in \mathbb{R}$ in terms of $x,\ y\ $ such that
$$A\cdot \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} a \\ b \end{bmatrix}\neq \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
and such that
$$A\cdot \begin{bmatrix} -b \\ a \end{bmatrix} = \begin{bmatrix} b \\ -a \end{bmatrix}.$$
I got this next equation system
$$\begin{cases} a x + b y = a\\ a y - b x = b\end{cases}$$
and finnaly I need to show that the transformation over $\mathbb{R^{2}}$ that given by multiplication by $A$ is the reflection through $span\left( \begin{bmatrix} a \\ b \end{bmatrix}\right)$ relative to $span\left( \begin{bmatrix} -b \\ a \end{bmatrix}\right)$
but I can't see how I can proceed from here. Would be glad for some help.
Since $x^2 + y^2 = 1$, we can write $x = \cos(\theta)$ and $y = \sin(\theta)$ for some $\theta \in [0, 2\pi)$. Consider the product of $A$ with an arbitrary unit vector $v = \begin{bmatrix} \cos(\alpha) \\ \sin(\alpha)\end{bmatrix}$, where $\alpha \in [0, 2\pi)$. Then: $$Av = \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta)\end{bmatrix}\begin{bmatrix} \cos(\alpha) \\ \sin(\alpha)\end{bmatrix} = \begin{bmatrix} \cos(\theta)\cos(\alpha) + \sin(\theta)\sin(\alpha) \\ \sin(\theta)\cos(\alpha) - \cos(\theta)\sin(\alpha)\end{bmatrix} = \begin{bmatrix} \cos(\theta - \alpha) \\ \sin(\theta - \alpha)\end{bmatrix}.$$ We aim to find the eigenvectors corresponding to $1$ and $-1$. That is, we need $\alpha$ so that we obtain either the same vector back, or the negative vector. Note that, if $v$ is an eigenvector (for $1$ or $-1$), then so too is $-v$. Changing from $v$ to $-v$ is equivalent to adding (or subtracting) $\pi$ from $\alpha$. So, by replacing $v$ with $-v$ as needed, we can even assume $\alpha \in [0, \pi)$.
When is $v$ an eigenvector corresponding to $1$? We require $\cos(\theta - \alpha) = \cos(\alpha)$ and $\sin(\theta - \alpha) = \sin(\alpha)$. This means that $\theta - \alpha = \alpha + 2k\pi$, or equivalently, $\theta = 2\alpha + 2k\pi$. for some integer $k$. As $\theta, 2\alpha \in [0, 2\pi)$ we must have $k = 0$, so $\alpha = \theta/2$. This gives us our first eigenvector:
$$v = \begin{bmatrix} \cos(\theta/2) \\ \sin(\theta/2)\end{bmatrix}.$$
I'll leave you to deduce an eigenvector for $-1$ (I think you'd have some good guesses).