Find the (exact) reflection of the vector v = (5, 1) across the line: y = 2x.
Hint: A sketch of v and the line may suggest an approach.
I found the matrix
-3/5 6/5
4/5 2/5
which seems like it gives the reflection across y=2x
But my question is: is there way to do this by rotating the axes and changing bases? (I'm pretty sure this is what the assignment is asking me to do)
I'm having a hard time visualizing it since no angle is given to put into the rotation equations for R2
i.e.
x'=xcosθ +ysinθ
y'=-xsinθ+ycosθ
Any help is greatly appreciated 🙂
Best Answer
The matrix $$ \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $$ is clearly a reflection in the $x$-axis. Therefore what we want to do is map $y=2x$ to the $x$-axis with a rotation, reflect, and then undo the rotation. Suppose our line has angle $\tan{\theta}$ to the $x$-axis (here, $\tan{\theta}=2$, from which we can find cosine and sine using $\cos^2+\sin^2=1$ and $\cos^2 = 1/(1+\tan^2)$, as $\cos{\theta}=1/\sqrt{5}$, $\sin{\theta}=2/\sqrt{5}$ ). Then a rotation that sends $y\cos{\theta}=x\sin{\theta}$ to $x=0$ is $$ \begin{pmatrix} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta} \end{pmatrix}. $$ Then the whole reflection is given by $$ \begin{pmatrix} \cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta} \end{pmatrix}, $$ which when you multiply out and simplify gives $$ \begin{pmatrix} \cos{2\theta} & \sin{2\theta} \\ \sin{2\theta} & -\cos{2\theta} \end{pmatrix} $$ Putting $\theta=\arctan{2}$ and re-expanding should give $$ \begin{pmatrix} -3/5 & 4/5 \\ 4/5 & 3/5 \end{pmatrix}.$$