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Thread: spanning set V

  1. #1
    n22
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    spanning set V

    Hello,
    How should I solve this question.thanks.

    Let X = {f,g,h} where
    f(x) = 3 + x, g(x) = 1 + x^5 and h(x) = 1 + x -2x^5
    Let V ={f is an element of F such that f(x)=α+βx+γx^5

    a)determine whether X is a spanning set for V.


    (a) Determine whether X is a spanning set for V .
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  2. #2
    MHF Contributor
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    Re: spanning set V

    Since $\displaystyle h(x)+2g(x) = 1+x-2x^5+2+2x^5 = 3+x = f(x)$, it can not be a spanning set. There are only two independent functions and the subspace has 3 dimensions.

    - Hollywood
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  3. #3
    MHF Contributor

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    Re: spanning set V

    Hollywood's answer is best. Being not as sharp, I would have gone directly to the definition of "span":

    Given any numbers, $\displaystyle \alpha$, $\displaystyle \beta$, and $\displaystyle \gamma$, do there exist numbers, a, b, and c, such that
    $\displaystyle a(3+ x)+ b(1+ x^5)+ c(1+ x- 2x^5)= \alpha+ \beta x+ \gamma x^5$
    $\displaystyle (3a+ b+ c)+ (a+ c)x+ (b- 2c)x^5= \alpha+ \beta x+ \gama x^5$

    So $\displaystyle 3a+ b+ c= \alpha$, $\displaystyle a+ c= \beta$, and $\displaystyle b- 2c= \gamma$
    From $\displaystyle a+ c= \beta$, $\displaystyle c= \beta- a$.
    From $\displaystyle b= \gamma+ 2c= \gamma+ 2\beta- 2a$.

    Putting those into the first equation, $\displaystyle 3a+ b+ c= 3a+ \gamm+ 2\beta- 2a+ \beta- a= \gamma+ 3\beta= \alpha$.

    Since the "a"s cancelled we cannot solve for a.

    In fact, in order that there exist such a representation of $\displaystyle \alpha + \beta x+ \gamma x^5$ we must have $\displaystyle \alpha= \gamma+ 3\beta$ so that $\displaystyle \alpha+ \beta x+ \gamma x^5= \gamma+ 3\beta+ \beta x+ \gamma x^5= (3+ x)\beta+ (1+ x^5)\gamma$. That is, $\displaystyle 3+x$, $\displaystyle 1+ x^5$, and $\displaystyle 1+ x- 2x^5$ span the two dimensional subspace having $\displaystyle \{3+ x, 1+ x^5\}$ as basis.
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