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Geometry of linear operator

  1. Jun 4, 2010 #1
    I have been re-reading some linear analysis the last week, and I have been playing around a bit with the linear subspace of [tex]\mathbb{R}^\mathbb{R}[/tex] that you get with the basis {sin x, cos x}. It didn't take me long to realize that the family of transformations parametrized by theta [tex]T_\theta : \sin(x) \mapsto \sin(x + \theta)[/tex] (cos likewise) is a family of linear transformations on the space. However, when I looked at their matrices I was a bit surprised. W.r.t. the basis above, it is given by

    [tex]T_\theta = \left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right][/tex]

    It can't be a coincidence that its matrix is a rotation matrix on [tex]\mathbb{R}^2[/tex]. Does anyone have a nice geometrical explanation of what is going on?
    Last edited: Jun 4, 2010
  2. jcsd
  3. Jun 4, 2010 #2


    Staff: Mentor

    For a given value of theta, the transformation T rotates a vector x by an angle of theta, measured in the counterclockwise direction.

    For example, if theta = pi/2,
    [tex]T_{\pi /2} = \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right][/tex]
    T_pi/2(<1, 0>^T) = <0, 1>^T.
  4. Jun 4, 2010 #3
    Mark44, T is not an operator on [tex]\mathbb{R}^2[/tex]. It is an operator on the function space spanned by sin x and cos x. To be more precise, any vector in the space that T is operating on is of the form a sin x + b cos x for a,b reals. T is not a rotation operator on that function space. What would that even mean?
    Last edited: Jun 4, 2010
  5. Jun 4, 2010 #4


    Staff: Mentor

    I skimmed over some of what you wrote in your OP, so I didn't catch some of the subtlety there. What do you mean "linear subspace of [tex]\mathbb{R}^\mathbb{R}[/tex]
    I'm not familiar with that notation, and don't see how it aligns with the transformation you described. As described, your transformation maps a one-dimensional space to a two-dimensional space. I don't see how you would get a 2 x 2 matrix out of this.
  6. Jun 4, 2010 #5
    [tex]\mathbb{R}^\mathbb{R}[/tex] is the set of all functions on [tex]\mathbb{R}[/tex], and is a vector space under scalar multiplication and addition of functions. sin x and cos x are members of this vector space and are linearly independent. They therefore form the basis of a subspace V. Any function in V can be written uniquely as a sin x + b cos x. The operator T sending sin x to sin(x + theta) and cos x to cos(x + theta) is actually a linear map on that space, and its matrix is as above with respect to the basis {sin x, cos x}. That is what I meant when I wrote "cos likewise" above - it operates on both basis functions linearly, and is a function from a two dimensional space into itself. Therefore its matrix is 2 by 2.

    For an example of another operator on that space, consider the differential operator
    [tex]D = \left[ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right] [/tex]

    As you point out, the matrix of T has the same form as a rotation matrix in [tex]\mathbb{R}^2[/tex]. It seems as if there should be a geometrical reason for that, and I'm not seeing it, I guess because I'm not sure how to go about thinking about the "geometry" of a space of functions.
    Last edited: Jun 4, 2010
  7. Aug 1, 2010 #6
    Let me start by some quick digression. The operator T acts on the functions by translation or shifting. In fact, T is a unitary operator (it preserves the L2 norm) If sin(x) and cos(x) were L^2 functions, we could easily see it. Unfortunately there are not. However if u take any 2 square integrable functions of R, you would be able to see that their norms are preserved under T. Better yet take a compact subset of the reals, and restrict sin and cos to it, compute their norms, and compute the L^2 norms of T(sin(x)) and T(cos(x)). They should be the same.

    Now back to your example. If you define an isomorphism between span{sin(x),cos(x)} and R^2, you could see that the translation action on span{sin(x),cos(x)} is simply equivalent to a rotation action on R^2. Here is the connection, T is norm preserving, thus the equivalent operator of T on R^2 should also be norm preserving. Here is a clue to the answer. What kind of operators on the plane are norm preserving operators? only rotation operators, translation operators or linear combination of both. Therefore, it should not be surprising that the action of T on R^2 is a rotation action of the plane.
    Vignon S. Oussa
    Last edited: Aug 2, 2010
  8. Aug 1, 2010 #7


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    Staff Emeritus
    Science Advisor
    Gold Member

    The space of all functions is irrelevant, since you only care about the algebra and geometry of the plane spanned by sin(x) and cos(x).

    (I'm pretty sure that) there are only two building blocks for how the circle group can act on a real vector space:
    • Doing nothing to a one-dimensional vector
    • Rotating a two-dimensional vector

    Since your transformations are not identity transformations, there has to be some basis where the operators are the standard rotation matrices.
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