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Using the standard basis of $\mathbb{R}^2$, determine the matrix of the following linear transformation

A reflection in a line forming an angle $\frac{\theta}{2}$ with the $x$-axis.

According to the solutions given in the book the result should be $\begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \\ \end{pmatrix}$.

I understand how to generally solve these types of problems, but the trigonometry is causing me difficulty. I would appreciate it if someone could include a graph and explain their workings in simple mathematical language.

Thank you.

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To find the matrix of the reflection $f$, it suffices to find $f(e_1)$ and $f(e_2)$, where $e_1=(1,0)$ and $e_2=(0,1)$. Both vectors have unit lengths and the same will be true for their images.

It might help if you draw a picture.

The vector $e_1$ is at angle $0$ and the image $f(e_1)$ will have angle $\theta$, so it is the vector $$f(e_1)=(\cos\theta,\sin\theta).$$

To find $f(e_2)$ the computation is slightly more complicated. Notice that $e_2$ has angle $\pi/2$ with the $x$-axis. And we have the axis of reflection, which is the line with the angle $\theta/2$. So the angle between $e_2$ and the axis of reflection is $\alpha=\pi/2-\theta/2$. And the angle of $f(e_2)$ will be $\pi/2-2\alpha$. (Notice that if we rotate $e_2$ by the angle $-\alpha$, we get a vector which is in the direction of the reflection axis. If we do it once again, it is the result of reflection.) So we need to calculate $\pi/2-2\alpha$.

Now it is some simple algebraic manipulation $$\frac\pi2-2\alpha = \frac\pi2-2\left(\frac\pi2-\frac\theta2\right) = \theta-\frac\pi2.$$ (As a kind of sanity check you might notice that $\frac{\frac\pi2+\left(\theta-\frac\pi2\right)}2=\frac\theta2$, so the reflection axis is exactly in the middle between $e_2$ and $f(e_2)$, just as expected.)

We know that the angle is $\theta-\pi/2$ and the vector has unit length, which means $$f(e_2)=\left(\cos\left(\theta-\frac\pi2\right),\sin\left(\theta-\frac\pi2\right)\right).$$

This can be further simplified. We have $\cos\left(\theta-\frac\pi2\right)=\cos\left(\frac\pi2-\theta\right)=\sin\theta$ and $\sin\left(\theta-\frac\pi2\right)=-\sin\left(\frac\pi2-\theta\right)= -\cos\theta$.

So we got $$f(e_2)=(\sin\theta,-\cos\theta).$$

And the matrix of the linear transformation is $$\begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \\ \end{pmatrix}.$$

The determinant is $$\begin{vmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \\ \end{vmatrix}=-\cos^2\theta-\sin^2\theta=-1.$$ (Exactly as expected, since this linear transformation preserves lengths but changes orientation.)

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If you decompose a given vector $\mathbf v$ into a component $\mathbf v_\parallel$ that’s parallel to the given line and a component $\mathbf v_\perp$ that is perpendicular to it, it should be reasonably clear from the following diagram that the reflection $\mathbf v'$ of $\mathbf v$ in this line can be found by reversing $\mathbf v_\perp$.

enter image description here

Symbolically, $\mathbf v'=\mathbf v_\parallel-\mathbf v_\perp$, but $\mathbf v_\perp = \mathbf v-\mathbf v_\parallel$, therefore $\mathbf v'=2\mathbf v_\parallel-\mathbf v$. So, the problem is reduced to finding $\mathbf v_\parallel$.

Let $\mathbf u=\langle\cos{\theta/2},\sin{\theta/2}\rangle^T$, i.e., a unit vector in the direction of the given line. Then $\mathbf v_\parallel$ is just the projection of $\mathbf v$ onto this vector, which is $$\begin{align}\mathbf v_\parallel &= (\mathbf u\cdot\mathbf v)\,\mathbf u \\ &= \left(v_x\cos{\theta\over2}+v_y\sin{\theta\over2}\right)\,\left\langle\cos{\theta\over2},\sin{\theta\over2}\right\rangle^T \\ &= \left\langle v_x\cos^2{\theta\over2}+v_y\cos{\theta\over2}\sin{\theta\over2}, v_x\cos{\theta\over2}\sin{\theta\over2}+v_y\sin^2{\theta\over2} \right\rangle^T,\end{align}$$ so $$\mathbf v' = \left\langle 2v_x\cos^2{\theta\over2}-v_x+2v_y\cos{\theta\over2}\sin{\theta\over2}, 2v_x\cos{\theta\over2}\sin{\theta\over2}+2v_y\sin^2{\theta\over2}-v_y \right\rangle^T.$$ In matrix form this is $$\begin{align}\mathbf v' &= \begin{bmatrix}2\cos^2{\theta\over 2}-1 & 2\cos{\theta\over2}\sin{\theta\over2} \\ 2\cos{\theta\over2}\sin{\theta\over2} & 2\sin^2{\theta\over2}-1\end{bmatrix}\mathbf v \\ &= \begin{bmatrix}\cos^2{\theta\over 2}-\sin^2{\theta\over 2} & 2\cos{\theta\over2}\sin{\theta\over2} \\ 2\cos{\theta\over2}\sin{\theta\over2} & \sin^2{\theta\over2}-\cos^2{\theta\over 2}\end{bmatrix}\mathbf v \\ &= \begin{bmatrix}\cos\theta & \sin\theta \\ \sin\theta & -\cos\theta\end{bmatrix}\mathbf v.\end{align}$$ The last step uses the double-angle formulas for sine and cosine.

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Try thinking of this transformation as a composition of three transformations. First, map the line across which we want to reflect to the $x$-axis. Then reflect across the $x$-axis, and finally rotate the $x$-axis back to the original line. The first rotation is given by the matrix \begin{equation} \left(\begin{matrix} \cos(-\theta/2) & -\sin(-\theta/2)\\ \sin(-\theta/2) & \cos(-\theta/2)\end{matrix}\right). \end{equation} (Are you familiar with rotation matrices?) Then the reflection across the $x$-axis is given by \begin{equation} \left(\begin{matrix} 1 & 0\\ 0 & -1\end{matrix}\right). \end{equation} Finally, we rotate back to our original line with \begin{equation} \left(\begin{matrix} \cos(\theta/2) & -\sin(\theta/2)\\ \sin(\theta/2) & \cos(\theta/2)\end{matrix}\right). \end{equation} We can obtain the desired matrix by multiplying these matrices together (pay attention to the order!).

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  • $\begingroup$ Thank you for the response. Are you sure your answer is correct? My solutions have the answer as the following: imgur.com/a/VN0Sr $\endgroup$ – The Pointer Sep 3 '16 at 19:22
  • $\begingroup$ Yeah, if you multiply the three given matrices in the correct order (and make the necessary simplifications), you'll get the matrix shown in your link. Given the simplicity of the final matrix, there may well be a more direct way to find it. $\endgroup$ – 211792 Sep 3 '16 at 19:30
  • $\begingroup$ Would the following answer be correct? f(e_1) = [ cos(theta/2), sin(theta/2)], f(e_2) = [cos(theta/2 + pi), sin(theta/2 + pi)] I've graphed it in the same way. $\endgroup$ – The Pointer Sep 3 '16 at 20:04
  • $\begingroup$ No, $f(e_1)$ and $f(e_2)$ will be the first and second columns, respectively, of the final matrix. If you take the three matrices above and write them right-to-left and multiply, you'll get the desired $2\times 2$ matrix. $\endgroup$ – 211792 Sep 3 '16 at 20:34
  • $\begingroup$ Sorry, but I still don't understand. :( The follow image is a crude drawing of my graph. The angle in quadrant 1 represents my first point f(e_1) and the angle in quadrant 3 represents my second point f(e_2). Is this correct or have I been doing this incorrectly? In terms of x = rcos(theta) and y = rsin(theta), the first angle is the vector [cos(theta/2), sin(theta/2)] and the second is [cos(theta/2 + pi), sin(theta/2 + pi)]? Please specify where I am going wrong. imgur.com/a/HNQYn $\endgroup$ – The Pointer Sep 3 '16 at 20:58

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