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Thread: Subspace of a polynomial

  1. #1
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    Subspace of a polynomial

    I'm having a lot of trouble with this topic. I have no idea on what to do for these questions:

    1) Show that the set $\displaystyle S=\{p\in\mathbb{P}_2(0)=1\}$ is NOT a subspace of $\displaystyle \mathbb{P}_2$

    The only thing I can think of doing for this question is making $\displaystyle p(x)=ax^2+bx+1$




    2) Show that the set:

    $\displaystyle S=\{p\in\mathbb{P}_3''(x)=0 \ \ \forall x \ \ \in\mathbb{R}\}$

    is a subspace of $\displaystyle \mathbb{P}_3$


    Would I do the following:

    $\displaystyle p(x)=ax^3+bx^2+cx+d$

    $\displaystyle \Longrightarrow p'(x)=3ax^2+2bx+c$

    $\displaystyle \Longrightarrow p''(x)=6ax+2b$

    $\displaystyle p''(x)=0 \ \mbox{for} \ x=-\dfrac{b}{3a}$
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  2. #2
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    Quote Originally Posted by acevipa View Post
    I'm having a lot of trouble with this topic. I have no idea on what to do for these questions:

    1) Show that the set $\displaystyle S=\{p\in\mathbb{P}_2(0)=1\}$ is NOT a subspace of $\displaystyle \mathbb{P}_2$

    The only thing I can think of doing for this question is making $\displaystyle p(x)=ax^2+bx+1$
    You want to show that the given set is not a subspace of $\displaystyle \mathbb{P}_2$. Recall the definition of a vector subspace. You need to show that one of the vector subspace axioms does not hold in this case; that is, you need to give a counter example to one of the axioms.


    2) Show that the set:

    $\displaystyle S=\{p\in\mathbb{P}_3''(x)=0 \ \ \forall x \ \ \in\mathbb{R}\}$

    is a subspace of $\displaystyle \mathbb{P}_3$


    Would I do the following:

    $\displaystyle p(x)=ax^3+bx^2+cx+d$

    $\displaystyle \Longrightarrow p'(x)=3ax^2+2bx+c$

    $\displaystyle \Longrightarrow p''(x)=6ax+2b$

    $\displaystyle p''(x)=0 \ \mbox{for} \ x=-\dfrac{b}{3a}$
    No. You need to show that the given set is a subspace of $\displaystyle \mathbb{P}_3$. Again, recall the axioms of vector space and show that they hold for any polynomials in your set S.

    I'll start with closure to addition: Let $\displaystyle p,q \in S$. Then $\displaystyle p''(x) = q''(x) = 0$ by definition of S. Now, note that $\displaystyle (p+q)'' = p'' + q'' = 0$, that is, $\displaystyle (p+q)'' = 0$ and so by definition of S, $\displaystyle (p+q) \in S$

    Can you do the rest?
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  3. #3
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    Quote Originally Posted by Defunkt View Post
    You want to show that the given set is not a subspace of $\displaystyle \mathbb{P}_2$. Recall the definition of a vector subspace. You need to show that one of the vector subspace axioms does not hold in this case; that is, you need to give a counter example to one of the axioms.




    No. You need to show that the given set is a subspace of $\displaystyle \mathbb{P}_3$. Again, recall the axioms of vector space and show that they hold for any polynomials in your set S.

    I'll start with closure to addition: Let $\displaystyle p,q \in S$. Then $\displaystyle p''(x) = q''(x) = 0$ by definition of S. Now, note that $\displaystyle (p+q)'' = p'' + q'' = 0$, that is, $\displaystyle (p+q)'' = 0$ and so by definition of S, $\displaystyle (p+q) \in S$

    Can you do the rest?
    For the first question, do you because $\displaystyle p(0)=1$ do we say that the zero vector is not included and therefore the set is empty.

    With the second question, how would you show that it is non-empty?
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  4. #4
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    Quote Originally Posted by acevipa View Post
    For the first question, do you because $\displaystyle p(0)=1$ do we say that the zero vector is not included and therefore the set is empty.
    Since $\displaystyle p(0)=1$ for any $\displaystyle p \in S$ then the zero polynomial is not in S. This, of course, does not mean that S is empty. But one of the subspace axioms is that the space must have the identity element. Here it does not, therefore S is not a subspace.

    With the second question, how would you show that it is non-empty?
    Can you think of any polynomial $\displaystyle p \in \mathbb{P}_3$ such that $\displaystyle p''(x) = 0 ~ \forall x \in \mathbb{R}$? (hint: its simple).
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  5. #5
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    Quote Originally Posted by Defunkt View Post
    Since $\displaystyle p(0)=1$ for any $\displaystyle p \in S$ then the zero polynomial is not in S. This, of course, does not mean that S is empty. But one of the subspace axioms is that the space must have the identity element. Here it does not, therefore S is not a subspace.



    Can you think of any polynomial $\displaystyle p \in \mathbb{P}_3$ such that $\displaystyle p''(x) = 0 ~ \forall x \in \mathbb{R}$? (hint: its simple).
    Okay, so would I just prove that a degree one polynomial $\displaystyle p(x)=ax+b$ is a subspace of $\displaystyle \mathbb{P}_3$?
    Last edited by acevipa; Sep 9th 2010 at 05:32 PM.
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