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Math Help - vector spaces & subspaces

  1. #1
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    vector spaces & subspaces

    Determine whether W is a subspace of V

    V=P_2
    W = \left\{ a + bx + cx^2 : abc=0 \right\}

    my solution:
    First step: W is nonempty because it contains the zero polynomial (a=b=c=0)
    Second step: Let
    p(x) = a + bx + cx^2
    and
    q(x) = d + ex + fx^2
    Then
    p(x) + q(x) = \left( a+d \right) + \left( b+e \right)x + \left( c + f \right)x^2

    So p(x) + q(x) is also in W (because it has the right form). Similarly, if k is a scalar, then

    kp(x) = ka + kbx + kcx^2

    so kp(x) is in W.

    Thus, W is a nonempty subset of p_2 that is closed under addition and scalar multiplication. therefore, W is a subspace of p_2

    Was just wondering if this was all good, becuase the initial condition of the set stated above (a*b*c=0) confused me a bit
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  2. #2
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    p(x) + q(x) will be in W if and only if (a+d)(b + e)(c + f) = 0. W contains all polynomials of second degree such that the product of the coefficients is equal to zero. Similiary, for kp(x) to be in W, we must have (ka)(kb)(kc) = 0. You have to check this before you can go on and say that p(x) + q(x) and kp(x) are in W.

    Hope this helps!
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  3. #3
    MHF Contributor

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    Quote Originally Posted by james12 View Post
    Determine whether W is a subspace of V

    V=P_2
    W = \left\{ a + bx + cx^2 : abc=0 \right\}

    my solution:
    First step: W is nonempty because it contains the zero polynomial (a=b=c=0)
    Second step: Let
    p(x) = a + bx + cx^2
    and
    q(x) = d + ex + fx^2
    Then
    p(x) + q(x) = \left( a+d \right) + \left( b+e \right)x + \left( c + f \right)x^2

    So p(x) + q(x) is also in W (because it has the right form).
    What do you mean "it has the right form"? What "form" are you talking about? Surely it is V because it is a quadratic but to be in W it must also have the product of its coefficients equal to 0 (which is the same as saying at least one of its coefficients is 0. Consider u= x^2+ 2, v= x^2+ 3x. Are they both in W? What is their sum? Is it in W?

    Similarly, if k is a scalar, then

    kp(x) = ka + kbx + kcx^2

    so kp(x) is in W.

    Thus, W is a nonempty subset of p_2 that is closed under addition and scalar multiplication. therefore, W is a subspace of p_2

    Was just wondering if this was all good, becuase the initial condition of the set stated above (a*b*c=0) confused me a bit
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  4. #4
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    Ahh it's all clear now, thanks for the help guys, much appreciated!!
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