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Change of Coordinate Matrix

  1. Apr 7, 2009 #1
    In geometry the change of variable,

    [tex]x = (2 / sqrt(5))x' - (1 / sqrt(5))y'[/tex] (#1)
    [tex]y = (1 / sqrt(5))x' + (2 / sqrt(5))y'[/tex] (#2)
    can be used to transform the equation [tex] 2x^2 - 4xy + 5y^2 = 1[/tex] into the simpler equation [tex](x')^2 + 6(y')^2 = 1[/tex], in which form it is easily seen to be the equation of an ellipse.

    [tex]B and B'[/tex] are the standard ordered basis and new rotated basis respectively

    My question:
    Why is BB' have such a representation with B and B'? Why wouldn't it be B'B?
  2. jcsd
  3. Apr 8, 2009 #2

    Can you try to state your question differently? I was following what you wrote up to that point, but I'm not familiar with the notation you're using at the end.
  4. Apr 9, 2009 #3
    So in this particular example,
    B' = { 1/sqrt(5)(2, 1), 1/sqrt(5)(-1, 2) }
    Question: what about the basis B, what is it's values?

    Geometrically the change of variable is (x,y) --> (x',y') is a change in the way that the position of a point P in the plane.

    The change of variable is actually a change from [tex][P]_B = (x, y)[/tex], the coordinate vector of the point P relative to the ordered basis B = {e1, e2}, to [tex][P]_B' = (x', y')[/tex], the coordinate vector of P relative to the new rotated basis B'.

    Notice also that the matrix
    | 2 -1 |​
    Q = 1/sqrt(5) | 1 2 |

    equals [tex]^B_B'[/tex], where I denotes the identity transformation on [tex]R^2[/tex]. Thus [tex][v]_B = Q[v]_B_'[/tex] for all v in [tex]R^2[/tex]. A similar result is true in general.

    Two questions for the last two sentences from above:
    1.) What exactly is [v]_B? What are the values for this vector, how is it obtained?
    2.) What do they mean by a similar result is true?


  5. Apr 9, 2009 #4

    Sorry, I still don't know what that notation means.

    But here's what I see is happening. The ellipse [tex]2x^2 -4xy + 5y^2 = 1[/tex] represents a quadratic form, meaning a particular inner product on R^2. Imagine that there is a symmetric matrix S with positive eigenvalues...

    S = \left( \begin{array}{cc} 2 & -2 \\ -2 & 5 \end{array} \right)

    Then we're looking at the set of all points [tex]v = \left( \begin{array}{c} x \\ y \end{array} \right)[/tex] satisfying [tex]Sv \cdot v = 1[/tex]. These form an ellipse. By finding the eigenvectors of that matrix S, and normalizing them, we assemble your rotation matrix Q. Q rotates the plane. Consider its transpose and see that [tex]v' = Q^Tv[/tex], and we get the ellipse [tex]x'^2 +6y'^2 = 1[/tex]. Note that these coefficients 1 and 6 are the eigenvalues of S, and they represent the maximum and minimum values of the ellipse's distance from the origin.

    So, what is v? This question I understand. What you are calling [tex]v_B[/tex]. Well, it can really be any point in the plane that you transform by [tex]Q^T[/tex], but in our case we can limit it to those points in the plane on the ellipse. That is, all points for which [tex]Sv \cdot v = 1[/tex].

    I don't understand your question about the values of the basis B. You wrote down the basis B.

    Like you, I also don't know what they mean by "a similar result is true".
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