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Area of a transformed region

  1. Jul 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Let T:R2→R2 be the linear transformation such that T(x,y)→(9x+3y,5x+2y). Let R be the region on the plane defined as {(x,y) [itex]\in[/itex] R2/ 0≤x≤2 and 0≤y≤2 }. Consider the region T(R) [itex]\subset[/itex] R2, which is the image of the region R by the linear transform T. What is the area of the region T(R)?

    PS: When I write purely R, I refer to the region. And when I write R2, to the plane. Just to make it clear
    2. Relevant equations
    N/A


    3. The attempt at a solution
    Well, I'm not making quite progress. I mean, I've notice some things, e.g. this linear transformation is a linear operator, that it is also surjective (it's image covers the whole plane). To begin with, I think the region R is like a little square, with area 4. But I get quite confused afterwards on what T(R) really looks like and then in computing it. Hope for some help, thanks in advance.
     
  2. jcsd
  3. Jul 14, 2014 #2
    If you are familiar with determinants you should use that. Otherwise you might want to try to draw the new region. Try to figure out what happens to the boundary. First take ##y=0## and let ## 0 \leq x \leq 2## and see where that line segment maps to. Then do the same for the other three edges of the square.
     
  4. Jul 14, 2014 #3

    HallsofIvy

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    Science Advisor

    Because, as you say, this is a linear equation, it maps straight lines to straight lines. The original figure is a square with vertices (0, 0), (2, 0), (0, 2), (2, 2). What are T(0, 0), T(2, 0), T(0, 2), and T(2, 2)?
     
  5. Jul 14, 2014 #4
    Thanks guys, I was able to figure out better how to do looking up what happened to the boudaries, indeed. Took T(0,0)=(0,0), T(2,0)=(18,10), T(0,2)=(6,4), T(2,2)=(24,14). The problem before was that I was doing exactly this, but was unable to realize that the transformed region is a parallelogram.
    I concluded that from calling u=(6,4), v=(18,10), w=(24,14). Since w-v=1u, w-v is actually parallel to u. Similarly, noticing w-u=1v, I could conclude that w-u was parallel to v. Afterwards, I took norm of the vector product between those u and v (considered them in R3, with z component 0), obtaining 12, which is numerically equal to the area (I hope!).
    But Quesadilla, could you tell me a little bit about this determinant approach? I'm not too bothered to work with it, might be interesting... Thanks again guys.
     
  6. Jul 14, 2014 #5
    Yes, so any linear transformation ##T : \mathbb{R}^n \rightarrow \mathbb{R}^n ##, ## \mathbf{x} \mapsto \mathbf{y}## can, in a given basis, be represented in matrix form
    \begin{equation*}
    \mathbf{y} = A\mathbf{x},
    \end{equation*}
    where ##\mathbf{x} = (x_1, x_2, ... x_n)##, ##\mathbf{y} = (y_1, y_2, ..., y_n) \in \mathbb{R}^n## are vectors and ##A \in \mathbb{R}^{n \times n}## is an ##n \times n## matrix.

    To any square matrix, one can associate a scalar value called the determinant, denoted ##det##. If you already know about determinants, you might know that a square matrix, ##A##, is invertible (i.e., there exists another matrix which we denote ##A^{-1}## such that ##A^{-1} A = A A^{-1} = I##. Here ##I## is the identity matrix.) if and only if ##det(A) \neq 0##.

    The value of the determinant also tells you the scaling factor an area (or volume in higher dimensions) is multiplied by under the linear transformation. The sign of the determinant (if it is nonzero) will also tell you whether orientation is preserved or reversed.

    In your case ##n = 2## and so the transformation can be written, in matrix form
    \begin{equation*}
    \begin{bmatrix}
    \tilde x \\ \tilde y
    \end{bmatrix}
    =
    \begin{bmatrix}
    9 & 3 \\ 5 & 2
    \end{bmatrix}
    \begin{bmatrix}
    x \\ y
    \end{bmatrix}.
    \end{equation*}
    There exist algorithms for determining the value of the determinant for any ##n##, but for ##2 \times 2## matrices it is very simple. If
    \begin{equation*}
    A = \begin{bmatrix} a & b \\ c & d \end{bmatrix},
    \end{equation*}
    the determinant is simply given by ## det(A) = ad - bc##.

    In your exercise, the determinant thus becomes ## 9 \cdot 2 - 3 \cdot 5 = 3##. As you noted ##R## is the square ##0 \leq x,y \leq 2## with area ##4##. The area of ##T(R)## is therefore ## 4 \cdot det(A) = 4 \cdot 3 = 12##, which agrees with what you found!

    If you want to know more about determinants you could read the article on wikipedia or any introductory linear algebra textbook.
     
  7. Jul 14, 2014 #6
    Hm, that's very nice to know about the scaling and orientation fact. My work with determinants back when I took linear algebra was more theoretical and I couldn't stand calculating them in higher order matrices. Always runned alway from it, hahaha. But in this case, I found this a more interesting way to go. Anyway, thanks for the help again!
     
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