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Find basis for vector space consisting of linear transformations

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Find a basis for V.
    Let W be a vector space of dimension 4. Let beta = {x1, x2, x3, x4 } be an ordered basis for W. Let V = {T in L(W) | T(x1) + T(x2) = T(x4) }


    2. Relevant equations
    L(W) is the set of linear transformations from W to W


    3. The attempt at a solution
    We know that V has dimension 3, since T(x4) can be expressed in terms of T(x1) and T(x2). So V has the general form of {T(x1), T(x2), T(x3), T(x1) + T(x2)}
    But I'm lost as to how I can find a basis for V when nothing is said explicitly.
     
  2. jcsd
  3. Dec 8, 2008 #2

    tiny-tim

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    Hi p3forlife! :smile:

    Hint: what is the matrix for a general T?

    what is the restriction on the matrix if T(x1) + T(x2) = T(x4)? :wink:
     
  4. Dec 8, 2008 #3
    So the matrix for a general T will look like:

    [T] = [a11 a12 a13 a14]
    [a21 a22 a23 a24]
    [a31 a32 a33 a34]
    [a41 a42 a43 a44]

    where a11 + a12 = a14
    a21 + a22 = a24
    a31 + a32 = a34
    a41 + a42 = a44

    but i can't get any farther than this unfortunately.
     
  5. Dec 8, 2008 #4

    tiny-tim

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    ok, let's do this slowly and logically …

    the basis for V must be smaller than any basis for L(W) …

    can you find a basis for L(W)? :smile:
     
  6. Dec 8, 2008 #5

    jambaugh

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    It may also help to use the linearity of your transformation [tex]T[/tex]. You should be able to rewrite your condition as [itex] T(u)=0[/itex] for some specific vector [itex]u[/itex].

    I am also not so sure of your dimension argument. Doesn't [tex]L(W)[/tex] start with dimension 16?

    Remember that (when acting on the left on a column vector) each row of a matrix is a linear functional mapping [tex]W \to [/tex] the real numbers.

    You can start with that simpler problem of finding a basis for the space of all linear functionals satisfying the given condition. Its answer should be very helpful in this problem.
     
  7. Dec 8, 2008 #6
    Okay, so finding a basis for L(W)...
    Since beta = { x1, x2, x3, x4} is a basis for W, if you do a linear transformation from W to W, the basis should be { T(x1), T(x2), T(x3), T(x4)} ?

    Sorry...I'm striking a blank about this problem.
     
  8. Dec 8, 2008 #7

    tiny-tim

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    Hi p3forlife! :smile:

    I don't think you understand what the elements of L(W) are …

    and a basis isn't attached to a particular T … it has to be the same basis for all T in L(W).

    How would you define a particular element of L(W)? :smile:
     
  9. Dec 8, 2008 #8
    So L(W) means you take an x in W, you apply a transformation, then you get T(x), where the set of all T(x) is the range of W.
     
  10. Dec 8, 2008 #9

    tiny-tim

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    ah, no …
    T (a typical member of L(W)) is the transformation. :smile:
     
  11. Dec 8, 2008 #10
    Argh....sorry this is taking so long :S

    So since L(W) is the set of all linear transformations from W to W, it means that V is a subset of a W, since there is the restriction of T(x1) + T(x2) = T(x4)
    In L(W), we start off with 16 dimensions, since each of T(xi) for i = 1, 2, 3, 4 has 4 elements, and 4 times 4 is 16. But, in V, since T(x1) + T(x2) = T(x4), we can take off 4 dimensions from 16, giving us 12 dimensions.

    So in order to satisfy T(x1) + T(x2) = T(x4),
    choose T(x1) = 1, T(x2) = T(x3) = 0, T(x4) = 1. So then the basis for V has to be in terms of x1, x2, x3, x4, so we have the matrix:

    x1 0 0 x1
    x2 0 0 x2
    x3 0 0 x3
    x4 0 0 x4
    0 x1 0 x1
    0 x2 0 x2
    0 x3 0 x3
    0 x4 0 x4
    0 0 x1 0
    0 0 x2 0
    0 0 x3 0
    0 0 x4 0

    Is this what you mean by a matrix for a general T?
     
  12. Dec 9, 2008 #11

    tiny-tim

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    Yes, but you haven't said why it's 4.

    You can't put x1 etc in your matrix …

    the elements of the matrix are numbers, and x1 is a vector. :frown:

    A general matrix is

    a11 a12 a13 a14
    a21 a22 a23 a24
    a31 a32 a33 a34
    a41 a42 a43 a44

    where the a's are all numbers,

    and the vector x1 + x2, for example, is

    1
    1
    0
    0

    So how would you write, using vectors and a matrix, T(x1) + T(x2) = T(x4)? :smile:
     
  13. Dec 9, 2008 #12

    Vid

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    L(W) has the same dimension as W.

    Hint: A transformation can be completely determined by what it does to the basis vectors.
     
  14. Dec 10, 2008 #13

    jambaugh

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    The way he described it L(W) does not have the same dimension as W. L(W) is not the dual space of functionals W* (which does have same dimension as W if it's finite)
    but rather [tex]L(W)=W\otimes W^*[/tex] which has the square of the dimension of W.
     
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