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Unable to show proposed transformation is nonlinear

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Show whether or not the following transformation T is a linear transformation, given the description of T:

    T maps each point in R2 w/ polar coords. (r, θ) to R2 w/ polar coords. (r, 2θ). T maps zero-element to itself (T(0) = 0)


    2. Relevant equations
    I suppose (x, y) = (rcos(θ), rsin(θ)) is relevant.


    3. The attempt at a solution

    I attempted to show one of the two properties of linear transformations did not hold. That is,

    T(x + y) = T(x) + T(y), where x,y are in R2
    and
    T(cx) = cT(x).

    I showed that T((r1cos(θ1), r1sin(θ1)) + (r2cos(θ2), r2sin(θ2))) = T(r1cos(θ1) + r2cos(θ2), r1sin(θ1) + r2sin(θ2)) = (r1cos(2θ1) + r2cos(2θ2), r1sin(θ1) + r2sin(2θ2)).

    This actually ends up equaling T(x) + T(y), when I believe it shouldn't.
    I encounter the same issue when trying to disprove T(cx) = cT(x).

    The only thing I can think of is that I cannot just multiply both my angles by 2 and call that the correct transformation. If I cannot do that, how am I supposed to multiply my angles? Can I coalesce the two cosines into one cosine and the two sines into one sine? I was not able to find a formula that allowed for that.

    Any help would be greatly appreciated. I do not want a solution so much as I want to see where I am going wrong.

    -Ian
     
  2. jcsd
  3. Oct 4, 2009 #2

    Office_Shredder

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    Too many variables in that attempt. If you want to show something is NOT a linear transformation, pick a very concrete example. Just guess for two vectors that aren't lined up and see if T(x+y)=T(x)+T(y). It usually helps to pick easy vectors with zeroes in some of their coordinates

    At the end of your "proof" as such you just doubled all the individual angles in the sines and cosines, but there's no real reason to believe that's the same as doubling the final angle of the vector you had
     
  4. Oct 4, 2009 #3
    Yes, your final sentence is what I know my problem is. I was sure of this while I was doubling the angles. The problem is I do not see how I would double the angles in a case where you are adding two elements.

    for T(x + y), I have no reason to assume that the elements have the same r or θ. So that means I have to add to get r1cos(θ1) + r2cos(θ2), r1sin(θ1) + r2sin(θ2) into a form where I can double the angles appropriately.

    Do you know a way to simplify the equation so I get such a form? I was able to find one formula, but it just turned into a multiplication of the cosines and sines, so that didn't really help me.

    I know I can just prove it by showing a case where the linear transformation won't work, but I still want to know how I can show T(x + y) != T(x) + T(y) in the more general case, just because I am interested.
     
  5. Oct 4, 2009 #4

    jbunniii

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    That's because

    [tex]T(r_1 \cos(\theta_1) + r_2 \cos(\theta_2), r_1 \sin(\theta_1) + r_2 \sin(\theta_2)) \neq (r_1 \cos(2\theta_1) + r_2 \cos(2\theta_2), r_1 \sin(2\theta_1) + r_2 \sin(2\theta_2))[/tex]

    i.e., as you suspected, you can't "double the angles" like this.

    I think you might find it easier to see the nonlinearity if the transformation can be written in rectangular coordinates. How can this be done? Let's start with polar coordinates and then convert.

    Our mapping is

    [tex]f(r \cos \theta, r \sin \theta) = (r \cos (2 \theta), r \sin (2 \theta))[/tex]

    But

    [tex]\cos(2\theta) = \cos^2 \theta - \sin^2 \theta[/tex]

    and

    [tex]\sin(2\theta) = 2 \sin \theta \cos \theta[/tex]

    so

    [tex]\begin{align*}f(r \cos \theta, r \sin \theta) &= (r(\cos^2 \theta - \sin^2 \theta), 2 r \sin \theta \cos \theta) \\
    &= \frac{(r^2 \cos^2 \theta - r^2 \sin^2 \theta, 2 r^2 \sin \theta \cos \theta)}{r} \end{align*}[/tex]

    Now just substitute [itex]x = r \cos \theta[/itex] and [itex]y = r \sin \theta[/itex] and simplify, to end up with

    [tex]f(x,y) = \frac{(x^2 - y^2, 2xy)}{\sqrt{x^2 + y^2}}[/tex]

    It should be pretty easy to check that this isn't linear.

    That is because this part is true! And it's pretty easy to check, either geometrically or by using the function I wrote above. Only additivity (T(x+y) = T(x)+T(y)) fails for this particular map.
     
  6. Oct 5, 2009 #5
    Thank you :). I was able to use the double-angle formula (which I didn't even think to use for some reason!) to show that property 1 is not satisfied. What I did; however, was allow the two values, r1 and r2 to be equal to 1. This should be valid as the property needs to hold for all r1 and r2, so I should be allowed to choose any one I like. Doing this allowed me to simplify the equation I came up with (similar, but not identical, to your equation) and to show in a straightforward manner that addition is not preserved.
     
  7. Oct 5, 2009 #6

    Office_Shredder

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    Still too hard.

    It's not linear over a set of basis vectors (picking (1,0) and (0,1) gives a fairly trivial proof), which means that it's pretty much never linear. Observe using only linear algebra: If x and y are linearly independent and T(x+y) = T(x)+T(y) then you've proven that T is in fact linear since T(cx)=cT(x) is true and we can write any vector as a linear combination of basis vectors (in this case our basis is x and y). So T(x+y)=T(x)+T(y) implies that x is a multiple of y and using T(cx)=cT(x) we know this always holds when y is a multiple of x.
     
  8. Oct 5, 2009 #7

    jbunniii

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    Yes, that's completely valid. In fact as others pointed out, you only need to find two points [itex]x[/itex] and [itex]y[/itex] for which [itex]T(x+y) \neq T(x) + T(y)[/itex]. Your solution finds infinitely many points, so you overachieved. However: mine found infinitely more! :approve:
     
  9. Oct 5, 2009 #8
    lol. well thank you for the help. ive used double angle a bunch of times, but for whatever reason i did not notice that i could use it to help me out in this problem.

    office_shredder, yeah I know I can use points where I know this will fail and show that as a proof. I included that for good measure :). I just wanted to go more in depth to see how the equation actually got modified and how that results in a nonlinear transformation. I usually do go about it the way you do, though. This time was an exception :).

    Thank you both of you!
     
  10. Oct 5, 2009 #9

    Office_Shredder

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    My post wasn't to provide evidence that you can just test two points, but to show that once you know it's non-linear, you can classify exactly when T(x+y)=T(x)+T(y) occurs using only linear algebra, since you said
    And my above post is a very elegant (I think) argument showing when this occurs. If you were really interested in just messing around with the sines and cosines that's entirely different of course :)
     
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