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The vector space of linear transformations

  1. May 28, 2013 #1
    Consider the operation of multiplying a vector in [itex]ℝ^{n}[/itex] by an [itex]m \times n [/itex] matrix A. This can be viewed as a linear transformation from [itex]ℝ^{n}[/itex] to [itex]ℝ^{m}[/itex]. Since matrices under matrix addition and multiplication by a scalar form a vector space, we can define a "vector space of linear transformations" from [itex]ℝ^{n}[/itex] to [itex]ℝ^{m}[/itex].

    My question is whether this connection between linear transformations and transformation matrices exists for linear mappings to and from arbitrary vector spaces. So given some general n-dimensional vector space U and m-dimensional vector space W, can every linear mapping from U to W be viewed as multiplication by a [itex]m \times n [/itex] transformation matrix ?

    Or is there a linear transformation which cannot be viewed as multiplication by a transformation matrix?

    BiP
     
    Last edited: May 28, 2013
  2. jcsd
  3. May 28, 2013 #2

    CompuChip

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    Yes, because you can always reduce your general situation down to ##\mathbb{R}^n##.

    Given a general n-dimensional vector space U, you can choose a basis ##\{ \hat e_1, \hat e_2, \ldots, \hat e_n \}## (and without loss of generality you can let it be orthogonal, for convenience). Now consider the map $$\phi: U \to \mathbb{R}^n, u_1 \hat e_1 + \cdots + u_n \hat e_n \mapsto ( u_1, \ldots, u_n)$$.
     
  4. May 28, 2013 #3
    What is [itex] \{ u_{1},u_{2}...,u_{n} \} [/itex] ?
    And once you've reduced to [itex]ℝ^{n}[/itex] and [itex]ℝ^{m}[/itex], is it necessarily the case that the linear transformation can be viewed as multiplication by a transformation matrix? Would we need to establish an isomorphism between them?


    BiP
     
  5. May 28, 2013 #4

    WannabeNewton

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    Every linear map between two finite dimensional vector spaces can be represented as a matrix between the associated euclidean spaces. This is called the matrix representation of the linear map and is a standard topic in most linear algebra textbooks. See, for example, chapter 5 of Lang "Linear Algebra".
     
  6. May 28, 2013 #5

    CompuChip

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    They are the components of the vector in U. In the definition of the function I wrote an arbitrary element of U as ##u_1 \hat e_1 + u_2 \hat e_2 + \cdots + u_n \hat e_n## which I can do because ##\{ \hat e_i \mid i = 1, \ldots, n \}## is a basis.

    To give you an idea: choose a basis ##\{ \hat e_1, \hat e_2, \ldots, \hat e_n \}## on ##\mathbb{R}^n## and ##\{ \hat f_1, \hat f_2, \ldots, \hat f_m \}## on ##\mathbb{R}^m##. Let ##T : \mathbb{R}^n \to \mathbb{R}^m## be a linear transformation. Then you can write
    $$T \hat e_i = a_{i,1} \hat f_1 + \cdots + a_{i,m} \hat f_m = \sum_{j = 1}^m a_{ij} \hat f_j$$
    The coefficients ##a_{ij}## form entries of an ##n \times m## matrix ##A##. It is straightforward to check that multiplying ##A## with a vector which contains a 1 in position i and 0 in all others (e.g. (0, 0, ..., 0, 1, 0, 0, ...)) gives you the correct coefficients, so that A is the matrix representation of T.
    Note that I make a distinction between the linear map T itself and the matrix A. Of course, you are free to choose another basis on ##\mathbb{R}^n## and/or ##\mathbb{R}^m##, which will change the numbers in the matrix. The transformation T doesn't change though - this can be confusing at first! :)

    Now if ##S: U \to V## is another transformation you can let ##\phi: U \to \mathbb{R}^n## be as I described earlier, mapping an element of U into a vector in ##\mathbb{R}^n## by picking out the components relative to some arbitrary basis. Let ##\psi: V \to \mathbb{R}^m## be a similar map for V. Now you can show that ##\phi## and ##\psi## are bijections and therefore for any u in U, you can write S(u) as ## \psi^{-1} A \phi u## for some suitable matrix A, which just represents a linear transformation ##\mathbb{R}^n \to \mathbb{R}^m##.

    If you don't follow the above paragraph, just think of it this way: if U is an n-dimensional vector space, you can see it as just ##\mathbb{R}^n## looking slightly different. So any map between U and an m-dimensional space V is secretly just a map from ##\mathbb{R}^n## to #\mathbb{R}^m##.

    (PS: Note that things do get trickier if the spaces are no longer finite-dimensional.)
     
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