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Math Help - Span and polynomials.

  1. #1
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    Span and polynomials.

    Question: Do the polynomials t^{3} + 2t + 1, t^{2} - t + 2, t^{3} +2, -t^{3} + t^{2} - 5t + 2 span P_{3}?

    My attempt: Let at^{3} + bt^{2} + ct + d be an arbitrary vector in P_{3}, then:
    c_{1}(t^{3} + 2t + 1) + c_{2}(t^{2} - t + 2) + c_{3}(t^{3} +2) + c_{4}(-t^{3} + t^{2} - 5t + 2) = at^{3} + bt^{2} + ct + d

    Combining the terms on the left side and equating corresponding coefficients gives the linear system:
    c_{1}           + c_{3}  - c_{4} = a
              c_{2}              + c_{4} = b
    2c_{1} - c_{2}           - 5c_{4} = c
    c_{1} + 2c_{2} + 2c_{3} + 2c_{4} = d

    Which becomes the augmented matrix:
    \left[\begin{array}{cccc|c}1&0&1&-1&a\\0&1&0&1&b\\2&-1&0&-5&c\\1&2&2&2&d\end{array}\right]
    Row reducing it, I come to the matrix:
    \left[\begin{array}{cccc|c}1&0&1&-1&a\\0&1&0&1&b\\0&0&-2&-2&c-2a+b\\0&0&0&0&-\frac{1}{2}(c-2a+b)(d-a-2b)\end{array}\right]

    Since there are a row of zeros, then -\frac{1}{2}(c-2a+b)(d-a-2b) must be equal to zero for the system to be consistent, and hence, the polynomials do not span P_{3}. However, my book concluded that the reason that the polynomials do not span P_{3} is because it is an inconsistent system, that no value for a, b, c, and d would yield 0 for the 4th row? How did they come to this conclusion? Thanks in advance.
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  2. #2
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    Re: Span and polynomials.

    my first instinct is to say check your arithmetic for your row-reduction

    EDIT: also, note that you are choosing a,b,c,d arbitrarily, you are not solving for them.
    Last edited by Deveno; October 9th 2011 at 06:54 PM.
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  3. #3
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    Re: Span and polynomials.

    Quote Originally Posted by Deveno View Post
    my first instinct is to say check your arithmetic for your row-reduction
    My row reductions:
    \left[\begin{array}{cccc|c}1&0&1&-1&a\\0&1&0&1&b\\2&-1&0&-5&c\\1&2&2&2&d\end{array}\right] \rightarrow \left[\begin{array}{cccc|c}1&0&1&-1&a\\0&1&0&1&b\\0&-1&-2&-3&c-2a\\0&2&1&3&d-a\end{array}\right] \rightarrow \left[\begin{array}{cccc|c}1&0&1&-1&a\\0&1&0&1&b\\0&0&-2&-2&c-2a+b\\0&0&1&1&d-a-2b\end{array}\right] \rightarrow \left[\begin{array}{cccc|c}1&0&1&-1&a\\0&1&0&1&b\\0&0&-2&-2&c-2a+b\\0&0&0&0&-\frac{1}{2}(c-2a+b)(d-a-2b)\end{array}\right]

    First row operations:
    r_{3} - 2r_{1} \rightarrow r_{3}
    r_{4} - r_{1} \rightarrow r_{4}
    Second row operations:
    r_{3} + r_{2} \rightarrow r_{3}
    r_{4} - 2r_{2} \rightarrow r_{4}
    Third row operation:
    r_{4} - \frac{1}{2}r_{3} \rightarrow r_{4}
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    Re: Span and polynomials.

    Quote Originally Posted by Deveno View Post
    EDIT: also, note that you are choosing a,b,c,d arbitrarily, you are not solving for them.
    Yes, that is why if I were to set each one of a, b, c, and d to 0, the system should be consistent. No?
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  5. #5
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    Re: Span and polynomials.

    the product on the last row in the last column should be a sum.

    if your four polynomials span P3, then one linear combination (at least) must be t^3 (which is a = 1, b = c = d = 0).

    this leads to 0 = -2 (from the 4th row, where your augmented column should read (d-a-2b)+(1/2)(c-2a+b)), which is inconsistent.

    the zero-vector (the polynomial 0t^3 + 0t^2 + 0t + 0 in this case) is ALWAYS in the span of ANY subset.

    thus setting a = b = c = d = 0 is a spectacularly bad choice for a test polynomial, because it tells us nothing (you can

    always make a linear combination of 0's be 0).
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