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Linear Transformation

  1. Sep 25, 2015 #1
    1. The problem statement, all variables and given/known data
    media%2F128%2F1286bc27-b87e-4d7f-8c5b-1b8ded2a0e2b%2FphpYSE3dF.png

    2. Relevant equations


    3. The attempt at a solution
    I would just like to know what is being requested when it asks me to draw sketches in order to illustrate that T is linear. Does it have something to do with altering to position of the line L itself? Any help would be very much appreciated. Thanks.
     
  2. jcsd
  3. Sep 25, 2015 #2

    Krylov

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    No, you keep the line ##L## fixed at all times in this exercise.
    1. Start by drawing a vector ##\mathbf{x}## in the diagram, say somewhere betwee the horizontal axis and the line ##L##. Now apply ##T## to ##\mathbf{x}##. Where does ##T(\mathbf{x})## go?
    2. After this, draw another vector ##\mathbf{y}## in the same way. Also draw ##T(\mathbf{y})## again.
    3. Next, add ##\mathbf{x}## and ##\mathbf{y}## using the parallelogram law. Call the result ##\mathbf{z}## and draw ##T(\mathbf{z})##.
    4. Finally, add ##T(\mathbf{x})## and ##T(\mathbf{y})## using the parallelogram law. If you have done the previous steps correctly, the resulting vector should coincide with ##T(\mathbf{z})##. Do you understand why this should be the case?
    The previous steps illustrate the additivity of ##T##. It remains to sketch the homogeneity of ##T##. You can probably figure out how?
     
  4. Sep 25, 2015 #3
    Thank you! Any suggestions on how to approach part b?
     
  5. Sep 25, 2015 #4

    Krylov

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    Hint: In your sketch in step 1. above, using rules from trigonometry, you should be able to express ##[T(\mathbf{x})]_1## and ##[T(\mathbf{x})]_2## in terms of ##x_1, x_2## and ##\theta##. The ##2 \times 2## matrix
    $$
    A =
    \begin{bmatrix}
    a_{11} & a_{12}\\
    a_{21} & a_{22}
    \end{bmatrix}
    $$
    you are asked to find, is such that
    $$
    \begin{bmatrix}
    [T(\mathbf{x})]_1\\
    [T(\mathbf{x})]_2
    \end{bmatrix}
    =
    A
    \begin{bmatrix}
    x_1\\
    x_2
    \end{bmatrix}
    $$
    This matrix is a function of ##\theta## alone.
     
  6. Sep 25, 2015 #5
    Thank you!
     
  7. Sep 25, 2015 #6

    HallsofIvy

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    Staff Emeritus
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    Note that if you apply matrix [itex]\begin{bmatrix}a & b \\ c & d\end{bmatrix}[/itex] to the vector [itex]\begin{bmatrix}1 \\ 0 \end{bmatrix}[/itex] the result is [itex]\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix}= \begin{bmatrix}a(1)+ b(0) \\ c(1)+ d(0)\end{bmatrix}= \begin{bmatrix}a \\ c \end{bmatrix}[/itex].

    That is, applying a matrix to the vector (1, 0) gives the first column of the matrix as a vector. Similarly, applying a matrix to the vector (0, 1) gives the second column of the matrix.

    Now, the vector (1, 0) is 10 degrees "below" the line of reflection so its reflection is 10 degrees "above" or 20 degrees above the x-axis. What are the components of a vector of length 1 that makes an angle 20 degrees above the x-axis? The vector (0, 1) is 80 degrees "above" the line of reflection so its reflection is 80 degrees "below" the line of reflection or 70 degrees below the x- axis. What are the components of a vector of length 1 that makes an angle 70 degrees below the x-axis?
     
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