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Linear transformations: function arguments

  1. Sep 2, 2014 #1
    I have a small confusion about functions and variables. So, on doing a bit of reading, a linear transformation is a function that maps inputs from one vector space to another.

    So, let us take for example a simple rotation matrix. This matrix takes a point in 2D space and maps it to another point in this 2D space. So, this is fine and the argument to this rotation function is a 2D point.

    Now, in many applications, we want to find the optimal rotation matrix i.e. we want to find the optimal angle [itex]\theta[/itex] to do the rotation. So, my question is can we view this rotation matrix now as a function of two variables i.e. [itex]\theta[/itex] and the input 2D point.

    All the explanations seem to treat the angle of rotation as a constant or in our optimisation problem case some unknown (but constant) quantity to estimate. So, in most optimisation cases it will involve taking the derivative of the rotation matrix wrt to the variable of interest i.e. [itex]\theta[/itex] and I was wondering whether I can view the rotation matrix as a function of [itex]\theta[/itex] i.e. write it as [itex]R(\theta, p)[/itex] where [itex]R[/itex] is the rotation function and [itex]p[/itex] is the point in 2D space.
  2. jcsd
  3. Sep 2, 2014 #2


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    As you said, most explanations treat the angle of rotation as an unknown constant, or parameter, which means that θ isn't considered to be a variable. If θ is allowed to vary (as in the optimization problems you cite), the rotation matrix is a function of only this rotation angle θ.
  4. Sep 2, 2014 #3
    Thanks for that explanation. As far as I understood, the optimisation is going (well, according to some local minima/maxima) to select the best rotation function which optimises the given cost function. So, in this case [itex]\theta[/itex] can be seen as a variable as we evaluate the cost function by varying [itex]\theta[/itex] continuously around the current value.
  5. Sep 2, 2014 #4


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    A function T from a vector space V to a vector space W also has to satisfy the condition ##T(ax+by)=aT(x)+bT(y)## for all x,y in V and all real numbers a,b.

    You can certainly define a function f by
    $$f(\theta,x,y)=\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}\begin{pmatrix} x\\ y\end{pmatrix},$$ but if you're looking for the optimal angle for a given (x,y), you might prefer to to define functions ##f_{x,y}## (one for each (x,y)) by
    $$f_{x,y}(\theta)=\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}\begin{pmatrix} x\\ y\end{pmatrix},$$
    Last edited: Sep 2, 2014
  6. Sep 4, 2014 #5
    It is a misconception that linear matrices/operators that represent such transformations as rotation and shear take points as arguments. They take and produce vectors. These are affine transformations (D × D+1 matrices) that take and produce points.
    Last edited: Sep 4, 2014
  7. Sep 4, 2014 #6


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    "Point" is a standard term for elements of (the underlying sets of) topological spaces. All finite-dimensional vector spaces can be given a topology in a simple way. Hilbert spaces are equipped with inner produces, which are used to define topologies in a standard way. So unless you're talking about some very exotic vector space, vectors are points.

    Some authors also use the term "point" to mean any element of any set. With that terminology, vectors are points even if you are talking about those exotic vector spaces.

    Pamparana was talking about ##\mathbb R^2##, which is just a plane with vector space operations defined on it. So even if we ignore the stuff I said about topology, it's still more than OK to call those vectors points, because they are points in the sense of Euclidean geometry.
  8. Sep 4, 2014 #7
    points or vectors?

    Didn’t spot the word “ℝ2”. Was my sight failed, or the browser maybe?

    Ῑ’m not willing to stuff the thread with off-topic. Just notice: “matrix takes a point” is a (formally correct, but) counter-intuitive and confusingly ungeometric terminology.
  9. Sep 4, 2014 #8


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    He said "point in 2D" space and then "2D point". So he was talking about ##\mathbb R^2## even though he never used that symbol.
  10. Sep 4, 2014 #9
    Fredrik: there is neither addition nor scalar multiplication for points on Euclidean plane, without an additional structure (such as distinguished point). ℝ2 defines a Euclidean plane, but an abstract Euclidean plane is neither ℝ2 nor a vector space.
  11. Sep 4, 2014 #10


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    Formally correct of course. But in the context, I think it's very clear that the OP meant ##\mathbb{R}^2##.
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