Deriving a 2D Rotation Matrix

Date: 11/17/2009 at 07:33:25
From: John
Subject: how to derive 2D rotation matrix geometrically

Using the following as an example, compute the point (x',y') where

  |x'| = |cos(theta)   -sin(theta)| |x|
  |y'|   |sin(theta)    cos(theta)| |y|

I watch a lot of physics lectures.  This comes up a lot in special 
relativity.  My question is the teacher tried to show how this matrix
was derived from the graph using geometry.  I got lost.  I have looked
at it over and over.  For the life of me I can't figure out how this
matrix is derived geometrically.  Can you help me with this?

John

Date: 11/17/2009 at 12:49:57
From: Doctor Tom
Subject: Re: how to derive 2D rotation matrix geometrically

Hi John,

I don't know if this will help, but I never remember the matrix and am
too lazy to look it up, so I re-derive it each time I need it.

If we know what the matrix does to the vectors (1,0) and (0,1), we've
got the whole thing.  (In your notation, those two vectors ought to be
written in columns, of course.)

So I draw the x-y axes, and then draw (1,0) rotated by theta.  It
obviously goes to:

  (cos theta, sin theta)

Similarly, drawing where (0, 1) goes on the axes, I see it goes to:

  (-cos theta, sin theta)

So what we want is a matrix such that:

  | a b |  |1| = | cos theta |
  | c d |  |0|   | sin theta |

And this forces a = cos theta and c = sin theta.

Doing the same thing for (0, 1) yields the other two values.

- Doctor Tom, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 11/18/2009 at 08:24:10
From: Doctor George
Subject: Re: how to derive 2D rotation matrix geometrically

Hi John,

Dr. Tom gave you a good explanation for how to derive the rotation
matrix.  There are a couple more ways to look at this.  Here is a link
to an article that shows another one:

  Rotation Matrix Using Trig Functions
    http://mathforum.org/library/drmath/view/70267.html 

Also, consider this.

  |x| = x |1| + y |0|
  |y|     |0|     |1|

In this representation, the vector (x,y) is a linear combination of
the coordinate axes unit vectors i and j.

To rotate the point, we just want to use a different set of unit vectors.

  |x'| = x |cos(theta)| + y |-sin(theta)|
  |y'|     |sin(theta)|     | cos(theta)|

Do you see that (cos(theta), sin(theta)) is the unit vector for the
rotated x axis, and (-sin(theta), cos(theta)) is the unit vector for
the rotated y axis?

Now we just rewrite the linear combination of the vectors as a matrix
multiplication.

  |x'| = |cos(theta)  -sin(theta)| |x|
  |y'|   |sin(theta)   cos(theta)| |y|

Does that make sense?  Write again if you need more help.

- Doctor George, The Math Forum
  http://mathforum.org/dr.math/