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Math Help - 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 (0)=1\}" alt="S=\{p\in\mathbb{P}_2(0)=1\}" /> is NOT a subspace of \mathbb{P}_2

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




    2) Show that the set:

    ''(x)=0 \ \ \forall x \ \ \in\mathbb{R}\}" alt="S=\{p\in\mathbb{P}_3''(x)=0 \ \ \forall x \ \ \in\mathbb{R}\}" />

    is a subspace of \mathbb{P}_3


    Would I do the following:

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

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

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

    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 (0)=1\}" alt="S=\{p\in\mathbb{P}_2(0)=1\}" /> is NOT a subspace of \mathbb{P}_2

    The only thing I can think of doing for this question is making p(x)=ax^2+bx+1
    You want to show that the given set is not a subspace of \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:

    ''(x)=0 \ \ \forall x \ \ \in\mathbb{R}\}" alt="S=\{p\in\mathbb{P}_3''(x)=0 \ \ \forall x \ \ \in\mathbb{R}\}" />

    is a subspace of \mathbb{P}_3


    Would I do the following:

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

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

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

    p''(x)=0 \ \mbox{for} \ x=-\dfrac{b}{3a}
    No. You need to show that the given set is a subspace of \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 p,q \in S. Then p''(x) = q''(x) = 0 by definition of S. Now, note that (p+q)'' = p'' + q'' = 0, that is, (p+q)'' = 0 and so by definition of S, (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 \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 \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 p,q \in S. Then p''(x) = q''(x) = 0 by definition of S. Now, note that (p+q)'' = p'' + q'' = 0, that is, (p+q)'' = 0 and so by definition of S, (p+q) \in S

    Can you do the rest?
    For the first question, do you because 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 p(0)=1 do we say that the zero vector is not included and therefore the set is empty.
    Since p(0)=1 for any 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 p \in \mathbb{P}_3 such that 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 p(0)=1 for any 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 p \in \mathbb{P}_3 such that p''(x) = 0 ~ \forall x \in \mathbb{R}? (hint: its simple).
    Okay, so would I just prove that a degree one polynomial p(x)=ax+b is a subspace of \mathbb{P}_3?
    Last edited by acevipa; September 9th 2010 at 05:32 PM.
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