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Area of figure, resulting from unit square transformation

  1. May 29, 2015 #1
    1- Let the linear operator on R^2 have the following matrix:


    A = 1 0
    -1 3

    What is the area of the figure that results from applying this transformation to the unit square?

    2- I am abit confused here, I thought that the matrix for the unit square would be,
    0 0
    1 0
    0 1
    1 1....
    But apparently that is not it.

    And after finding the unit square matrix, I blv I would find the images of it by computing A^Tx

    I am not sure though
     
  2. jcsd
  3. May 29, 2015 #2

    DEvens

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    A square would be four points. You seem to have collected four points that make up one particular unit square, and arranged them as a 2x4 array. That is kind of odd.

    What does it mean to be a linear operator on R^2? What does the operator A operate on? Hint: R^2 means there are two real numbers involved. But there seem to be 8 real numbers involved in your square. So how is that going to work?
     
  4. May 29, 2015 #3
    oh. ok, so are they asking to find the Ax? where x would be the unit square.. so like
    [10. ] [ 1 0]
    [ 1 3 ] [ 01 ] ...
     
    Last edited: May 29, 2015
  5. May 30, 2015 #4
    anyone?
     
  6. Jun 1, 2015 #5

    HallsofIvy

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    Staff Emeritus
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    That matrix is NOT the unit square. The unit square is a geometric figure, not a matrix, that has vertices at (0, 0), (1, 0), (1, 1), and (0, 1).
    The bottom side, from (0, 0) to (1, 0), the vector [itex]\begin{bmatrix}1 \\ 0 \end{bmatrix}[/itex], is mapped into
    [tex]\begin{bmatrix}1 & 0 \\ -1 & 3\end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}1 \\ -1\end{bmatrix}[/tex]

    The left side, from (0, 0) to (0, 1), the vector [itex]\begin{bmatrix}0 \\ 1 \end{bmatrix}[/itex], is mapped into
    [tex]\begin{bmatrix}1 & 0 \\ -1 & 3 \end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix}= \begin{bmatrix}0 \\ 3\end{bmatrix}[/tex]

    The other two sides get mapped into equivalent vectors so this is a parallelogram with vertices at (0, 0), (1, -1), (1, 2), and (0, 3).
    It should be easy to find the area of that parallelogram.

    (And, it is worth noting that this matrix has determinant 3.)
     
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