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Help! Linear Algebra transformations, is my understanding correct?

  1. Aug 8, 2010 #1
    Hello,

    I would like to check if my understanding of this linear algebra problem dealing with transformations is correct:

    vovcp4.png

    Part (1)

    we have the following coordinates systems:

    [tex] \tilde{x} = \begin{pmatrix} x_1 \\ x_2 \\ 1 \end{pmatrix} [/tex] and [tex] x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} [/tex]

    and the following equation : [tex] \tilde{x}^T \tilde{A} \tilde{x} = 0 [/tex]

    This parts only asks for getting the A matrix (symmetric matrix with coefficients) from a quadratic curve. Taking a look at the final parts of the problem, I guess this part is some sort of coordinates change from [tex] \tilde{x} \rightarrow x [/tex] using a matrix A, is this right? (like some sort of 3D-2D transformation)


    Part (2)

    we have the following coordinates systems:

    [tex] x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} [/tex] and [tex] y = \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} [/tex]

    and the following equation : [tex] x = Py [/tex]

    My thoughts: The equation in (i) corresponds to a quadratic curve, which has been rotated (it has a cross product term) and translated (it has terms y1^2, y1). In this part we want to transform (i) into (ii). In this transformation we undo the rotation (by rotating) and the matrix used is for it is P, transforming x coords into y coords.


    Part (3)

    we have the following coordinates systems:

    [tex] y = \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} [/tex] and [tex] z = \begin{pmatrix} z_1 \\ z_2 \end{pmatrix} [/tex]

    My thoughts: The equation in (ii) corresponds to a quadratic curve, which has been translated (it only has terms y1^2, y1). In this part we want to transform (ii) into (iii). In this transformation we undo the translation (by translating), with an equation like this:

    [tex] z = y + M [/tex]

    transforming coords y into coords z.


    Part (4.A)

    we have the following coordinates systems:

    [tex] z = \begin{pmatrix} z_1 \\ z_2 \end{pmatrix} [/tex] and [tex] \tilde{z} = \begin{pmatrix} z_1 \\ z_2 \\ 1 \end{pmatrix} [/tex]

    and the following equation : [tex] \tilde{z}^T \tilde{B} \tilde{z} = 0 [/tex]

    This parts looks like part 1, getting the matrix from a quadratic curve. Taking a look at the rest of the parts of the problem, I guess this part is a coordinates change from [tex] z \rightarrow \tilde{z} [/tex] using a matrix B, is this right?


    Part (4.B)

    we have the following coordinates systems:

    [tex] \tilde{z} = \begin{pmatrix} z_1 \\ z_2 \\ 1 \end{pmatrix} [/tex] and [tex] \tilde{x} = \begin{pmatrix} x_1 \\ x_2 \\ 1 \end{pmatrix} [/tex]

    and the following equation : [tex] \tilde{x} = \tilde{Q} \tilde{z} [/tex]

    This part is the composite transformation from (i), using coords [tex] \tilde{x} [/tex] to (iii) using coords [tex] \tilde{z} [/tex]. So composing all transformations from the previous parts into one, my guess is that :

    [tex] \tilde{Q} = ((\tilde{A} P) + M)\tilde{B} [/tex]

    would this be ok?

    Thanks in advance and sorry for the long post...
     
  2. jcsd
  3. Aug 8, 2010 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You want
    [tex]\begin{bmatrix}x_1 & x_2 & 1\end{bmatrix}\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ 1\end{bmatrix}= x_1^2+ x_2^2- 6x_1x_2+ 2x_1+ 4x_2+ \frac{15}{9}[/tex]
    That gives you a number of equations for a, b, c, d, e, f, g, h, and i.


     
  4. Aug 8, 2010 #3
    Thanks for your reply!

    Yes... I already got the values for a-i. What are you trying to say ?
     
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