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We have that:

The linear transformation matrix for a reflection across the line $y = mx$ is:

$$\frac{1}{1 + m^2}\begin{pmatrix}1-m^2&2m\\2m&m^2-1\end{pmatrix} $$

So reflection across the x-axis would be: $$\frac{1}{1 + (0)^2}\begin{pmatrix}1-(0)^2&2(0)\\2(0)&(0)^2-1\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0&-1\end{pmatrix}$$

But I’m having trouble thinking through what the matrix would be when we want to reflect every vector across the y-axis...

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3 Answers 3

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First of all, reflection with respect to $y$ axis cannot be handled by formula :

$$S=\frac{1}{1 + m^2}\begin{pmatrix}1-m^2&2m\\2m&m^2-1\end{pmatrix}\tag{1}$$

for the reason that, exceptionally, there is no $m$ for which the $y$ axis has equation $y=mx$ (one could consider that $m \to \infty$ but it would be necessary to invoke a continuity of the symmetry operator...).

Happily, there is an easy way to bypass this difficulty. Here is how.

Let us remark that, in general, $y=mx$ can be written $y=(\tan \alpha) \ x$ where $\alpha$ is the (polar) angle made by the straight line with $x$ axis. Therefore we can identify :

$$m=\tan(\alpha)\tag{2}$$

If we plug (2) into (1), we recognize the so-called "tangent half angle formulas" (https://www.math24.net/weierstrass-substitution/) :

$$S=\begin{pmatrix}\cos(2 \alpha)&\ \ \ \sin(2 \alpha)\\ \sin(2 \alpha)&-\cos(2 \alpha)\end{pmatrix}\tag{3}$$

which has strong similarities with a rotation matrix : it is an isometry matrix (its columns are unit length and are orthogonal). The difference with a rotation matrix is that its determinant is $-1$, attesting that a symmetry reverses figures' orientation).

As a consequence, if one looks for symmetry with respect to $y$ axis, corresponding to an angle $\alpha=\tfrac{\pi}{2}$, plugging this value into (3) gives matrix :

$$S=\begin{pmatrix}\cos(\pi)&\ \ \ \sin(\pi)\\ \sin(\pi)&-\cos(\pi)\end{pmatrix}=\begin{pmatrix}-1&0\\ \ \ 0&1\end{pmatrix}\tag{4}$$

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Hint: Just take the limit $m\to \pm \infty$.

So, you get the matrix $\begin{pmatrix}-1&0\\0&1\end{pmatrix}$.

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You only need the point on the plane with different-sign $x$-entry:

$$\begin{pmatrix}-1&0\\0&1\end{pmatrix}\implies \begin{pmatrix}-1&0\\0&1\end{pmatrix}\binom xy=\binom{-x}y$$

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