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Homework Help: Inverse mapping theorem , Transformations

  1. Nov 30, 2007 #1
    A quick question this time...

    Example: Let (u,v)=f(x,y)=(x-2y, 2x-y).
    Find the region in the xy-plane that is mapped to the triangle with vertices (0,0),(-1,2),(2,1) in the uv-plane.

    (0,0)=f(0,0), (-1,2) = f(5/3,4/3), and (2,1)=f(0,-1), the region is the triangle with these vertices.

    My question is:
    Yes, we get three points, but how do you know FOR SURE that the region is a TRIANGLE? I am lost here...

    Thanks for explaining!
  2. jcsd
  3. Nov 30, 2007 #2


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    You know "for sure" because the transformation is LINEAR- it maps straight lines into straight lines. If that's not enough, see what the transformation does to the line between (0,0) and (-1, 2) in the uv-plane. That is, of course, v= -2u.

    Since u= x-2y and v= 2x-y, that becomes 2x-y= -2(x- 2y)= -2x+ 4y. Then adding y and 2x to both sides we have 4x= 5y or y= (4/5)x, the equation of a straight line.

    Do the same with the other two sides of the triangle, v= (1/2)u and v= -(1/3)u+ 5/3 to see that they are mapped into straight lines.
  4. Nov 30, 2007 #3
    Is there any quick way to see that this transfromation is linear?

    "it maps straight lines into straight lines" <---is this always true for linear transfomations and is it what a linear transformation means geometrically? (I was never aware of the geometrical meaning of a linear transformation, I was just given the definition in my linear algebra course)
    Last edited: Nov 30, 2007
  5. Nov 30, 2007 #4


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    How can you be dealing with a problem like this and not know what a "linear" function is?
    f(x,y)=(x-2y, 2x-y) is linear because it involves only sums and differences of products of numbers with a variable. There are no powers of variables, products of different variables, or more complicated functions. Notice that you could also write this as a single matrix multiplication:
    [tex]f(x,y)= \left[\begin{array}{cc} 1 & -2 \\ 2 & -1\end{array}\right]\left[\begin{array}{c} x \\ y\end{array}\right][/tex]
    That's a sure sign of a linear transformation.

    Yes, it is fairly easy to prove that a linear transformation maps lines into lines.
  6. Nov 30, 2007 #5
    But the definition of linear transformation that I've learnt is:

    T: U->V is a linear transformation iff T(au+bv)=aT(u)+bT(v) for all u E U, v E V, for all a, b E R
  7. Nov 30, 2007 #6


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    and a matrix multiplying vectors in Rn satisfies that perfectly well
  8. Dec 1, 2007 #7
    but by seeing that x-2y, 2x-y are linear polynominals, is it enough to say that f(x,y)=(x-2y, 2x-y) is a linear transformation? (these 2 definitions of "linear" seem quite distinct to me...for example, a linear polynominal allows a constant term, but a linear transfomration does not...)
    Last edited: Dec 1, 2007
  9. Dec 1, 2007 #8


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    Yes, but seeing that x-2y, 2x-y are linear polynomials without constant term is enough!
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