# Simple Transformation

1. May 3, 2008

### Hevonen

[SOLVED] Simple Transformation

I have a simple reflection illustraded in matric form that should be presented in geometric form:

// cos(5pi/6) sin(5pi/6) //
// sin(5pi/6) -cos(5pi/6) //

(this should look like 2*2 matrix; see the file "transformation.jpg" if you want to see it prettier)

SOLUTION
"This matrix represents a reflection in y = x*tan(5pi/12)".

However, I do not understand why the angle is now 5pi/12 and not 5pi/6 as in the matrix. Moreover, what is the difference if the minus sign was in the front of the other cos( ) in the left hand corner and not in the right hand corner? How would the angle differ? Please, help me.

#### Attached Files:

• ###### transformation.JPG
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Last edited: May 3, 2008
2. May 3, 2008

### uart

I think you're getting confused because it's a reflection instead of a rotation. With a rotation, the angle of the rotation is equal to the angle of the cos and sin in the transformation (though the matrix is somewhat different for a rotation). In a reflection the angle of the cos and sin is twice the angle of the reflection axis. You probably have to draw a picture of it and work it through to really explain it properly (which I'm too lazy to do right now).

3. May 3, 2008

### Hevonen

I totally agree with you. The problem is that I do not know how to visualise the angle in reflection with different matrices. I only know that the angle is somehow halved in reflection when writing from matrix to geometric form.

4. May 3, 2008

### flebbyman

The reflection matrix when reflecting over a line, making angle x with x axis is:

// cos(2x) -sin(2x) //
// sin(2x) cos(2x) //

Thats why the "angle size" doubles

5. May 3, 2008

### Hevonen

I thought that rotation is of the form:

// cos(x) -sin(x) //
// sin(x) cos(x) //

and then reflection would be of the form:

// -sin(2x) cos(2x) //
// cos(2x) sin(2x) //

Have I understood something wrong? I think there is no "direction" in reflection as in rotation so is this form equivalent to yours:

// cos(2x) -sin(2x) //
// sin(2x) cos(2x) //

Last edited: May 3, 2008
6. May 3, 2008

### HallsofIvy

Staff Emeritus
Think of it as first transforming to the line of reflection, then to the final figure. Since the beginning and ending figures are symmetric about the line of reflection, the angle between each and that line is half the angle between them.

7. May 4, 2008

### Hevonen

Thank you. I can now visualise the angle. But are these matrices equivalent?

// -sin(2x) cos(2x) //
// cos(2x) sin(2x) //

// cos(2x) -sin(2x) //
// sin(2x) cos(2x) //
(I have seen this presentation of reflection in some exercises)

The difference is that the other is multiplied by a M matrix:
M =
// 1 0 //
// 0 1 //

But I do not understand what this M transformation do to the reflection. Is it still the same reflection?

8. May 4, 2008

### trambolin

Take a triangle and put it in such a way that it's bottom is sitting on the x axis. What I mean by Reflection with respect to the x axis, is keep everything same except the y coordinate of each point is negated i.e. I have $\vartriangle \to \triangledown$. Then, suppose the points are given as vectors where elements of the vectors are the (x,y) components. The only thing that I should do is multiply each point by the matrix,

$$\left( {\begin{array}{c} x_{new} \\ y_{new} \\ \end{array}} \right)= \left[ {\begin{array}{cc} 1 & 0 \\ 0 & { - 1} \\ \end{array}} \right] \left( {\begin{array}{c} x \\ y \\ \end{array}} \right)$$

If you consider the angle here, what do you think? What if I rotate pi radians around the origin? Then regarding your last question what if I rotate -pi radians ? After having all those, I think you can figure out what to do with reflection with respect to any given axis.

Last edited: May 4, 2008
9. May 4, 2008

### Hevonen

I have to correct my writing. I meant:

$$\left [ {\begin{array}{cc} -sin(2x) & cos(2x) \\ cos(2x) & sin(2x) \\ \end{array}} \right] = \left [ {\begin{array}{cc} cos(2x) & -sin(2x) \\ sin(2x) & cos(2x) \\ \end{array}} \right] * \left [ {\begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array}} \right]$$
EQUATION ENDS

What is the purpose of the matrix:
$$\left [ {\begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array}} \right]$$
Can it be shown geometrically?

Last edited: May 4, 2008
10. May 4, 2008

### Hevonen

Rotating -pi radians only changes the direction of rotation. And I think that matrix $$\left [ {\begin{array}{cc} -1 & 0 \\ 0 & 1 \\ \end{array}} \right]$$ would reflect the shape with respect to the y-axis.

I still do not know what is the function of the matrix $$M = \left [ {\begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array}} \right]$$ in the reflection?

Last edited: May 4, 2008
11. May 4, 2008

### trambolin

Well, now I am confused how to answer but let me try this... what if I write like this,

[whatever multiplies this matrix take the second component] = [2nd component]
[whatever multiplies this matrix take the first component ] [1st component ]

it was before

[1st] ==> [2nd]
[2nd] ==> [1st ]

I don't quite get what your problem is but can you reset your question and pose it again ?

12. May 5, 2008

### Hevonen

I will recheck my exercises. It may be that this thread already target my problem. Thanks!

One small question: is this matrix B used in reflection? It may be that "doubling the angle" is equivalent to use the matrix B :)

$$B= \left [ {\begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array}} \right]$$

Last edited: May 5, 2008
13. May 5, 2008

### Hevonen

My question is in the file where there is the exercise and the solution that has confused me.

#### Attached Files:

• ###### Question about simple transformation1.jpg
File size:
28.8 KB
Views:
57
14. May 5, 2008

### Hevonen

The same file in JPG.

15. May 5, 2008

### trambolin

16. May 5, 2008

### trambolin

17. May 5, 2008

### Hevonen

Thanks! Btw, is there available any program that could visualise transformations? I have Maple but I do not know whether it can do transformations. I think geometric presentation is much more efficient way of learning these :)