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Find the standard matrix of the transformation $T:\mathbb{R}^2\to \mathbb{R}^2$ that corresponds to the reflection through the line $x_2=2x_1$ followed by reflection through the line $x_1=3x_2$

I am very confused about this question, could anyone explain how to solve the problem? THANKS!

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First you change coordinates to where your basis is given by $\begin{pmatrix} 1 \\ 2\end{pmatrix}$ and $\begin{pmatrix} 2 \\ -1\end{pmatrix}$. This is done by multiplying by $$\begin{pmatrix} 1 & 2 \\ 2 & -1\end{pmatrix}^{-1} = \frac{1}{5}\begin{pmatrix} 1 & 2 \\ 2 & -1\end{pmatrix}.$$ Next, you reflect the second coordinate. This is done by multiplying by $$ \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}.$$ Finally, you change your coordinates back by multiplying by $$\begin{pmatrix} 1 & 2 \\ 2 & -1\end{pmatrix}.$$ Thus, your transformation is multiplication by $$ \frac{1}{5}\begin{pmatrix} 1 & 2 \\ 2 & -1\end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix} 1 & 2 \\ 2 & -1\end{pmatrix} = \frac{1}{5}\begin{pmatrix} 1 & 2 \\ 2 & -1\end{pmatrix}\begin{pmatrix} 1 & 2 \\ -2 & 1\end{pmatrix} = \frac{1}{5}\begin{pmatrix}-3 & 4 \\ 4 & 3 \end{pmatrix}.$$

Now try to do the second reflection yourself.

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  • $\begingroup$ You should explain to OP where the 1/5 factor comes from, as the determinant of the matrix. $\endgroup$ – Pieter Geerkens Apr 20 '13 at 13:04
  • $\begingroup$ I assumed that the OP knows how to invert $2\times 2$ matrices. If not, the result can at least be checked relatively easy I suppose. $\endgroup$ – Abel Apr 20 '13 at 13:06
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    $\begingroup$ If the OP infers that the general process of inverting a 2x2 matrix is necessary for this problem, your solution has missed the mark. $\endgroup$ – Pieter Geerkens Apr 20 '13 at 13:12
  • $\begingroup$ @PieterGeerkens, the OP in this question, just as in any other question, is assumed to be a responsible student who won't accept blindly anything written here: he, and not Abel, must check the hints (or more than hints) given to him and verify everythings fine. What he infers or not is his/her problem, and if he has any further doubt he can always write back or ask a new question. $\endgroup$ – DonAntonio Apr 20 '13 at 13:45
  • $\begingroup$ @DonAntonio: Answers are written for future readers as well as the current OP. With a slight enhancement, I believe Abel will have a significantly better answer, and possibly get more up votes (including mine). $\endgroup$ – Pieter Geerkens Apr 20 '13 at 13:52
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One method found in most linear algebra books to do reflections easily is to use a Householder transformation.

It says that if you have a unit column vector $u$ which is normal to your line of reflection, then the matrix for reflection across that line is $I_2-2uu^T$. If we label this matrix $X$, then you would multiply column vectors on the right of this matrix to see the result of the transformation.

For your first reflection, you can find that a unit normal is $u=\begin{bmatrix}\frac{-2}{\sqrt{5}}\\\frac{1}{\sqrt{5}}\end{bmatrix}$. (If this is unclear, than you should graph the line $x_2=2x_1$ in the $x_1 x_2$ plane. You'll have to do soemthing similar to figure out the normal to $x_2=\frac{x_1}{3}$.)

Then computing the rest:

$$uu^T=\begin{bmatrix}\frac{-2}{\sqrt{5}}\\\frac{1}{\sqrt{5}}\end{bmatrix}\begin{bmatrix}\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{bmatrix}=\begin{bmatrix}\frac{4}{5}&\frac{-2}{5}\\\frac{-2}{5}&\frac{1}{5}\end{bmatrix}$$

$$ 2uu^T=\begin{bmatrix}\frac{8}{5}&\frac{-4}{5}\\\frac{-4}{5}&\frac{2}{5}\end{bmatrix} $$

$$ I_2-2uu^T=\begin{bmatrix}\frac{-3}{5}&\frac{4}{5}\\\frac{4}{5}&\frac{3}{5}\end{bmatrix} $$

That would accomplish the first reflection! Now: can you find a unit normal for the second line, and carry out the same process? Once you do, you will have a new matrix $Y$ for the next reflection. Taking our old reflection $X$, the composition of the two reflections will be $YX$.


I'm no good at making or including images here yet, but if you do a google image search for "Householder reflection," you'll find a variety of diagrams which illustrate why the $I_2-2uu^T$ formula works.


Another potential solution is to use the fact that the product of two reflections is a rotation. That being the case, you could just focus on where a non-origin point lands after it is reflected. In fact, you could pick the point to be on the axis of the first reflection (so that it doesn't move) and then compute where the second reflection takes it. After you know where it lands, measure the angle $\theta$ between where it was and where it landed, and the rotation would be given by $\begin{bmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{bmatrix}$

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  • $\begingroup$ Made edits to correct an unfortunate misstep early on. Thanks for being patient! $\endgroup$ – rschwieb Apr 20 '13 at 13:37

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