I am asked to prove that an isometric mirroring on a line l through the origin on a 2 dimensional plane of the reals, has a determinant of -1.
First of all, is it true that my book is sloppy here in assuming unstatedly that such a mirroring is given by a matrix? because a determinant is only defined for a matrix right? and there could be some other computational implementation of such a mirroring that doesn't use matrices. Or am I missing something?
Moreover, can I simply assume that the transformation function that gives this mirroring must be of the form f(v) = Av, where A is a transformation matrix?
lastly, how does one go about proving this, even assuming that this is indeed the functional form?
Let $y=mx$ be the line, we wanna show several things. First, the map of mirroring is a linear map: this statement is actually much easier to see if we consider the basis $\vec v_1=(1,m),\vec v_2=(-m,1)$ that consists of perpendicular vectors. Then the map is just reflection along the given line (on which $\vec v_1$ lies), so it must be linear since we are just taking a component ($\vec v_2$) of a vector to its inverse ($-\vec v_2$), which is a linear combination of vectors. Specifically for any vector written as: $a_1 \vec v_1 + a_2 \vec v_2$,
$$T( a_1 \vec v_1 + a_2 \vec v_2)= a_1 \vec v_1 - a_2 \vec v_2$$
Then we want to show that the determinant can only be $-1$: consider the linear map as a matrix, thus should have matrix of the form $\begin{pmatrix} 1& 0 \\ 0 &-1\end{pmatrix}$ when respect to the basis. Since determinant is unchanged under change of basis, so it can only be $-1$.