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Understanding Theorem 4.2.5

  1. Apr 8, 2015 #1
    Theorem 4.2.5 The most general linear map ## f : R^3 → R ## is of the form ## x → x·a##, for some vector a in ##R^3##.
    Proof: Suppose that ##f : R^3 → R## is linear, and let ##a = (a_1, a_2, a_3)##, where ##a_1 =
    f (i), a_2 = f (j), a_3 = f (k).##
    Then ##f (x) = f (x_1i + x_2j + x_3k) = x_1 f (i) + x_2 f (j) + x_3 f (k) = x·a##.
    My understanding
    The function, f is scalar product and it takes two vectors x and a and changes them into a scalar x? why x again are they different? How did ## f(x) ## become ##f (x) = f (x_1i + x_2j + x_3k) ##?
     
  2. jcsd
  3. Apr 8, 2015 #2

    Mark44

    Staff: Mentor

    No, f maps a vector ##\vec{x}## to the scalar ##\vec{x} \cdot \vec{a}##. The dot (or scalar) product produces a scalar as its output.
    ##f(\vec{i}) = \vec{i} \cdot \vec{a} = a_1##. Similar for ##f(\vec{j})## and ##f(\vec{k})##.
     
  4. Apr 8, 2015 #3

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    I think s/he is making use of some representation theorem for functionals in
    ## \mathbb R^n## in which every linear functional can be described as the inner-product by a fixed element.
     
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