Linear Transformation in terms of Polar Coord.

aredian
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Homework Statement


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.

Homework 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

The Attempt at a Solution


I know x1 & x2 in terms of polar coordinates is
x1 = r cos a
x2 = r sin a

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

Thanks
 
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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) ] ?
 
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.
 
Defennder said:
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.

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
 
is it a counter clock wise rotation by a, to whatever the line is?
 
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.
 
Dick said:
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.

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

You guys are the best! Thank you very much!
 
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