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Thread: Answer check (Vector space)

  1. #1
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    Answer check (Vector space)

    Hi, can someone just check my answer, please?

    Question:
    Two subspaces S and T of $\displaystyle P_2(\mathbb{R})$ are given by

    $\displaystyle S = \lbrace f(x) ~| ~3f(0)+f'(0) = 0 \rbrace $
    $\displaystyle T = \lbrace f(x) ~| ~f(1) = 0 \rbrace$.

    Obtain a non-trivial quadratic $\displaystyle g = ax^2 + bx +c$ such that $\displaystyle g \in S\cap T$.

    My reponse:
    Let $\displaystyle g = ax^2 + bx +c \in P_2(\mathbb{R})$,
    Then $\displaystyle g' = 2ax + b$.

    For membership in S: $\displaystyle 3c + b=0 ~ (1)$.
    For membership in T: $\displaystyle a+b+c=0 ~ (2)$.

    From (1), $\displaystyle b=-3c$.
    From (2), $\displaystyle a=-b-c$.

    Let c = 1.
    $\displaystyle \therefore b=-3$ and $\displaystyle a=3-1=2$.

    So $\displaystyle g=2x^2-3x+1 \in S\cap T$ is one such non-trivial quadratic.
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  2. #2
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    Quote Originally Posted by scorpion007 View Post
    Hi, can someone just check my answer, please?

    Question:
    Two subspaces S and T of $\displaystyle P_2(\mathbb{R})$ are given by

    $\displaystyle S = \lbrace f(x) ~| ~3f(0)+f'(0) = 0 \rbrace $
    $\displaystyle T = \lbrace f(x) ~| ~f(1) = 0 \rbrace$.

    Obtain a non-trivial quadratic $\displaystyle g = ax^2 + bx +c$ such that $\displaystyle g \in S\cap T$.

    My reponse:
    Let $\displaystyle g = ax^2 + bx +c \in P_2(\mathbb{R})$,
    Then $\displaystyle g' = 2ax + b$.

    For membership in S: $\displaystyle 3c + b=0 ~ (1)$.
    For membership in T: $\displaystyle a+b+c=0 ~ (2)$.

    From (1), $\displaystyle b=-3c$.
    From (2), $\displaystyle a=-b-c$.

    Let c = 1.
    $\displaystyle \therefore b=-3$ and $\displaystyle a=3-1=2$.

    So $\displaystyle g=2x^2-3x+1 \in S\cap T$ is one such non-trivial quadratic.
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