Linear Transformation in terms of Polar Coord.

1. Jun 18, 2008

aredian

1. The problem statement, all variables and given/known data
Let L(x) be the Linear operator in R^2 defined by
L(x) = (x1 cos a - x2 sin a, x1 sin a + x2 cos a)^T

Express x1, x2 & L(x) in terms of Polar coordinates.
Describe geometrically the effects of the L.T.

2. Relevant equations
Well I know that:
a = tan^-1 (x2 / x1)
r = (x1^2 + x2^2)^1/2

where a is the angle in both cases

3. The attempt at a solution
I know x1 & x2 in terms of polar coordinates is
x1 = r cos a
x2 = r sin a

But Im not sure about L(x)... cos a & sin a stay the same in both coord?

Thanks

2. Jun 18, 2008

aredian

If they stay the same which I guess so... then by the properties of trigonometric functions...
L(x) in polar coord will be
[ (r cos 2a) , (r sin 2a) ] ?

3. Jun 18, 2008

Defennder

Yeah that looks ok. Now observe what are the differences in the original vector in polar coordinates and the transformed vector. That should help geometrically.

4. Jun 18, 2008

aredian

well... Obiously the a was doubled. But if the polar coordinates are of the form (r, a), How should I graph ( r cos 2a , r sin 2a )?

:S

5. Jun 18, 2008

aredian

is it a counter clock wise rotation by a, to whatever the line is?

6. Jun 18, 2008

Dick

I really don't think you are supposed to take the angle the same for both x and L. L is a matrix which expresses a rotation by an angle a. If a general point p=(r*cos(t),r*sin(t)) then L(p) will rotate t->t+a, or maybe t->t-a, I'll let you figure out which.

7. Jun 18, 2008

aredian

In my notes I have L(x)=(r*cos(a+t) , r*sin(a+t)). I though they would be the same so I reduced it to 2*a.

So I can graph a point P as you described it, anywhere and then increase the angle by a to prove my point? No need to figure out where exactly r*cos(a) or r*sin(a) will actually be located?

8. Jun 18, 2008

Dick

I don't think so. Just draw a picture that shows a linear transformation that rotates a vector by an angle a.

9. Jun 18, 2008

aredian

You guys are the best! Thank you very much!!!!