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Math Help - Find a basis for teh corresponding cyclic subspace

  1. #1
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    Find a basis for teh corresponding cyclic subspace

    For each of the following find a basis for the corresponding cyclic subspace:

    a) T = d/dx : R<=3[x] --> R<=3[x], v = x2

    b) T = d/dx: R<=3[x] --> R<=3[x], v = x3

    c) T is rotation of R2 by 180 degrees, v = [3,4]

    d) T is rotation by 30 degrees, v = [3,4]

    e) T: R3 --> R3 given by T(x,y,z) = (x+2z, 2x-y, z), and v = [1,0,0]


    My thoughts on what bases are:

    a) Would a basis be (x^2, 2x, 2)?
    b) Would a basis be (x^3, 3x^2, 6x)?
    c) Would a basis be {(3,4),(-3,-4)}?
    d) Would a basis be {(3,4),(-3,-4)}? *Not too sure on this one*
    e) I have no idea ??
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Find a basis for teh corresponding cyclic subspace

    By R_{<=3}[x] do you mean the space of deg(3) or less polynomials with real coefficiants?
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    Re: Find a basis for teh corresponding cyclic subspace

    Yes you are correct jakncoke.
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    Senior Member jakncoke's Avatar
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    Re: Find a basis for teh corresponding cyclic subspace

    b and d seem to be incorrect. recall for d, the transformation matrix of rotation is given by  \begin{bmatrix} cos \theta & - sin \theta \\ sin \theta & cos \theta \end{bmatrix} For the last one, transformation matrix is given by  \begin{bmatrix} 1 & 0 & 2 \\ 2 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
    Last edited by jakncoke; February 20th 2013 at 06:02 PM.
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    Re: Find a basis for teh corresponding cyclic subspace

    So for b) would I have to go a step further and say the basis is {3x^2, 6x, 6}? or am I missing something between a and b that makes a correct and b incorrect?
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  6. #6
    Senior Member jakncoke's Avatar
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    Re: Find a basis for teh corresponding cyclic subspace

    the basis i got for b) was { x^3.3x^2. 6x.6}
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  7. #7
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    Re: Find a basis for teh corresponding cyclic subspace

    For e) I am having trouble coming up with a basis. Do I multiply the given vector by the transformation matrix, or is the basis just {(1,0,2),(2,-1,0),(0,0,1)}?
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  8. #8
    Senior Member jakncoke's Avatar
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    Re: Find a basis for teh corresponding cyclic subspace

    the basis is  (1,0,0) , (1,2,0)
    so you have \{(1,0,0), Tv, T^2v\} we see that Tv = (1,2,0) and  T^2v = (1,0,0)
    so  \{(1,0,0), (1,2,0), (1,0,0) since the third vector makes this linearly dependent, throw it away
    and thus resulting \{(1,0,0), (1,2,0) is the basis
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