# Linear Algebra Problem

1. Jun 19, 2005

### Mathman23

Rotating a figure around its center

Hi

I have this here this here linear transformation:

$$(x,y) \rightarrow \left[ \begin{array}{cc} -cos( 2 \theta ) & sin(2 \theta) \\ sin(2 \theta) & cos(2 \theta) \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right]$$

Turns the figure recthangel A(3,1) B(5/2, 2) , (9/2, 3) , D(5,2)

90 degrees clockwise from the figures current position.

My questions is how do I change the transformation so it turns the figure around the figures center ???

Do I add the figures center like this

$$(x,y) \rightarrow \left[ \begin{array}{cc} -cos( 2 \theta ) & sin(2 \theta) \\ sin(2 \theta) & cos(2 \theta) \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] + \left[ \begin{array}{c} a \\ b \end{array} \right]$$

Fred

Last edited: Jun 19, 2005
2. Jun 19, 2005

### HallsofIvy

Staff Emeritus
Yes, but first translate the rectangangle so its center is the origin, then rotate, then translate back. That is: determine what the coordinates of the center are, subtract those coordinates from each of the corners, multiply by the matrix then add the coordinates back.

3. Jun 19, 2005

### Mathman23

Hi,

I get the center of the figure to be (15/4,2)

Then I subtract the coordinants from the center:

A(3,1): there is A'((3/4,1) is the new coordinant.

I then insert this coordinant into my transformation

$$(x,y) \rightarrow \left[ \begin{array}{cc} -cos( 2 \theta ) & sin(2 \theta) \\ sin(2 \theta) & cos(2 \theta) \end{array} \right] \left[ \begin{array}{c} \frac{3}{4} \\ 1 \end{array} \right]$$

where theta = pi/2

I get the rotated A to be [(3/4,-1)]

I add the diffence between old A rotated A' to obtain the correct rotated A coordinant??

If I do that I get the center again. That can't be the rotated koordinant can it ?

Sincerely

Fred

Last edited: Jun 19, 2005
4. Jun 19, 2005

### OlderDan

The center should return to its original position. That is why you shift it to the origin before rotating, and then shift it back. Perform the same steps on the corners of the rectangle and see what you get.

5. Jun 19, 2005

### Mathman23

This is the coordinants translated:

A(3,1) -------> (3/4,-1)

B(5/2,2) -----> (5/4,0)

C(9/2,3) -----> ((-3/4,1)

D(5,2) -------> (-5/4,1)

Is this what You mean ?

How does this prove that my proposed transformation is correct ??

/Fred

6. Jun 19, 2005

### OlderDan

If you plot them you will see that these are not corect. I did not notice it earlier because I thought you were asking about the center point, but your translation A to A' was not correct. You failed to change the y coordinate (or maybe just got the sign wrong?) and the sign of the x coordinate is wrong. Be careful with the translation and try these again.

7. Jun 19, 2005

### Mathman23

Hi Dan,

As mentioned earlier my assignment is write a transformation which rotates the rectangel 90 degrees clockwise from its current position around the figures center.

My surgestion for the transformation is that wrong?

You talk about that I need to translate the center from to the (0,0) and then back again to the original center in order to make the transformation valid. Can this be incorparated into my surgested transformation if its vallid ??

I have two follow up questions if thats allright.

$$s(x) = \left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]$$

I'm told that this transformation can be done by another transformation using homogeneous coordinants.

$$s(x) = \left[ \begin{array}{ccc} a_{11} & a_{12} & 0 \\ a_{21} & a_{22} & 0 \\ 0 & 0 & 1 \end{array} \right]$$

Any idears on how I can show this ??

Finally how can one explain using transformations that a circle in the plane can be mapped into a ellipse ??

I have tried for three weeks to answer these three questions, but my Linear Algebra isn't very good and I'm in a terrible jam because I have to be able to answer these three questions in tuesday. Therefore I will be forever greatful if You could help me answer these question :-)

Best Regards and Bless You

Fred