Linear transformations

**SyNtHeSiS** · October 9th, 2010, 06:41 AM

Find the 2 x 2 matrix that represents a reflection in the line y = xtanalpha followed by a reflection in the line y = xtanbeta, where alpha, beta are constants. What is the linear map represented by this matrix?

I dont know what to do. I can only see that tanalpha = y/x and tanbeta = y/x therefore tanalpha = tanbeta, but this doesnt help.

**HallsofIvy** · October 9th, 2010, 07:09 AM

The line $y= tan(\alpha) x$ is a straight line through the origin at angle $\theta$ to the x-axis.

One way of determining a matrix representing any linear transformation is to see what it does to the basis vectors.
To find the reflection of the point (1, 0) in the line $y= tan(\alpha x= mx$ (for simplicity, I am going to write $m= tan(\alpha)$ ) we look at the line through (1, 0) perpendicular to the line y= mx.
The line through (1, 0) perpendicular to y= mx has slope -1/m and so equation y= -(1/m)(x- 1).

It also makes a right triangle with hypotenuse from (0, 0) to (1, 0) and angle $\alpha$ and so the "opposite side", the length of the line from (1, 0) to the original line, of $sin(\alpha)$ . The reflection of (1, 0) in the line y= mx, is at that same distance on the other side of the line. That is, it lies on the line y= -(1/m)(x- 1) and has distance $2sin(\alpha)$ from (1, 0) equal to $2sin(\alpha)$ . Using the distance formula we want $\sqrt{(x- 1)^2+ y^2}= \sqrt{(x- 1)^2+ (-(1/m)(x- 1)^2}= |x-1|\sqrt{1+ 1/m^2}= |x- 1|\sqrt{m^2+ 1}/m= 2sin(\alpha)$ .

Now, use the fact that $m= tan(\alpha)$ . $m^2+ 1= tan^2(\alpha)+ 1= sec^2(\alpha)$ so that equation becomes $|x- 1|\frac{sec(\alpha)}{tan(\alpha)}= \frac{|x- 1|}{sin(\alpha)}= 2\sin(\alpha)$ so [tex]|x- 1|= 2sin^2(\alpha)[tex]. It is clear, geometrically, that x< 1 so |x- 1|= 1- x and we have $1- x= 2sin^2(\alpha)$ and $x= 1- 2sin^2(\alpha)= 1- sin^2(\alpha)- sin^2(\alpha)= cos^2(\alpha)- sin^2(\alpha)= cos(2\alpha)$ . From y= -(1/m)(x- 1), $y= -(1/tan(\alpha))(-2sin^2(\alpha))= 2sin(\alpha)cos(\alpha)= sin(2\alpha)$ .

That is, the first reflection maps $(1, 0)$ to $(cos(2\alpha), sin(2\alpha)$ and that means that the first column of the matrix corresponding to that reflection is $\begin{bmatrix}cos(2\alpha) \\ sin(2\alpha)\end{bmatrix}$ .

There is probably some simpler formula to get that, but that is my take on the situation.

Do the same kind of analysis with (0, 1) to get the second column of the matrix.

Of course, the matrix representing the second reflection will be exactly the same thing with $\beta$ instead of $\alphs$ . The matrix representing both reflections, one after the other, is the product of those two matrices (in the correct order).

**SyNtHeSiS** · October 9th, 2010, 09:51 AM

Thanks for the post. It helped a lot.

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