$R$ is a reflection in the line $y=2x$.
$S$ is a rotation clockwise through angle $\theta$ .
Find a condition on $\theta$ for which the line $ y = x$ is an invariant line of $S^{-1}RS$
My working so far:
$R$ is a reflection in the line $y=\tan(\alpha)x$ so $2 = \tan(\alpha) \implies \alpha= \tan^{-1}(2)$
Let $a=\tan^{-1}(2)$
$R = \begin{pmatrix}\cos(2a)&\sin(2a)\\\sin(2a)&-\cos(2a)\end{pmatrix}$
$S=\begin{pmatrix}\cos(\theta)&\sin(\theta)\\-\sin(\theta)&\cos(\theta)\end{pmatrix}$
Starting with any point on the line $y=x$. The image after the combined transformation should remain on the line because it is an invariant line?
$S^{-1}RS\begin{pmatrix}x\\x\end{pmatrix}=\begin{pmatrix}x^"\\x^"\end{pmatrix}$
Multiplying by $S$
$RS\begin{pmatrix}x\\x\end{pmatrix}=S\begin{pmatrix}x^"\\x^"\end{pmatrix}$
After this I'm not sure how to find $\theta$
I've tried to calculate the combined transformation and then solve for $\theta $ but that hasn't worked. What could I do next to find $\theta$? Any help with this would be appreciated. Thanks.
Best Answer
Start by reducing the number of unknowns. If $\tan\alpha=2$, then $\cos\alpha=\pm1/\sqrt5$ and $\sin\alpha=\pm2/\sqrt5$. Expanding $R$ using standard double-angle formulas, you then have $$R = \begin{bmatrix}\cos^2\alpha-\sin^2\alpha & 2\cos\alpha\sin\alpha \\ 2\cos\alpha\sin\alpha & \sin^2\alpha-\cos^2\alpha\end{bmatrix} = \begin{bmatrix}-\frac35 & \frac45 \\ \frac45 & \frac35\end{bmatrix}$$ for either choice of signs.
Next, for a line through the origin to be invariant under some linear transformation $L$, it must be the case that for any $\mathbf v$ on the line, $L(\mathbf v)=\lambda v$ for some scalar $\lambda\ne0$. Moreover, because of linearity of $L$, this scalar $\lambda$ is independent of $\mathbf v$ (verify this!). So, instead $S^{-1}RS(x,x)^T = (x'',x'')^T$, write the right-hand side as $\lambda(x,x)^T$. Since this equation must hold for all $x$, it must hold for, say, $x=5$. This gives you a system of equations in $\lambda$ and $\theta$ that shouldn’t be too hard to solve.
Alternatively, we have $$S^{-1}RS = \frac15\begin{bmatrix}-3\cos{2\theta}-4\sin{2\theta} & 4\cos{2\theta}-3\sin{2\theta} \\ 4\cos{2\theta}-3\sin{2\theta} & 3\cos{2\theta}+4\sin{2\theta}\end{bmatrix}.$$ Observe that $$\frac1{5^2}\left((3\cos{2\theta}+4\sin{2\theta})^2+(4\cos{2\theta}-3\sin{2\theta})^2\right) = 1,$$ so $S^{-1}RS$ is itself a reflection. Now, the only invariant lines of a reflection in a line through the origin are the line itself and its perpendicular through the origin. Since we want the line $y=x$ to be invariant, there are only two possibilities for this reflection: $$S^{-1}RS = \begin{bmatrix}0&\pm1\\\pm1&0\end{bmatrix}.$$ Setting this equal to the other expression for $S^{-1}RS$ obtained previously produces the system of equations $$4\cos{2\theta}-3\sin{2\theta}=\pm5 \\ 3\cos{2\theta}+4\sin{2\theta} = 0,$$ which is equivalent to $\tan{2\theta}=-\frac34$, with the quadrant determined by the sign in the first equation of the system.