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Thread: Polynomial vector spaces

  1. #1
    Member Mollier's Avatar
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    Polynomial vector spaces

    Hi,

    problem:
    Consider the vector space $\displaystyle \mathbb{P}$ and the subsets $\displaystyle \mathbb{V}\;of\;\mathbb{P}$ consisting of those vectors (polynomials) $\displaystyle x$ for which

    (a) x has degree 3,

    (b) 2x(0)=x(1)

    (c) $\displaystyle x(t)\geq0$ whenever $\displaystyle 0\leq t\leq 1$,

    (d) $\displaystyle x(t)=x(1-t)$ for all $\displaystyle t$.

    In which of these cases is $\displaystyle \mathbb{V}$ a vector space?

    attempt:
    I first assume that $\displaystyle \mathbb{P}$ is a vector space over $\displaystyle \mathbb{C}$ as this is not mentioned in the problem.

    (a) Not a vector space.
    $\displaystyle x(t)=a_3t^3+a_2t^2+a_1t+a_0$
    Closed under vector addition.
    Closed under scalar multiplication.
    It does not have an origin that is a polynomial of degree 3 such that $\displaystyle x(t)+0(t)=x(t)$.

    (b) Is a vector space.
    $\displaystyle x(t) = a_nt^n+\cdots+a_1t+(a_n+\cdots+a_1) $
    Closed under vector addition.
    Closed under scalar multiplication.
    There exists an element (origin) $\displaystyle 0(t)$, such that $\displaystyle x(t)+0(t)=x(t)$
    $\displaystyle 2\cdot0(0)=0(1)$

    (c) and (d), could need a push in the right direction here.

    Thanks!
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  2. #2
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    Quote Originally Posted by Mollier View Post
    Hi,

    problem:
    Consider the vector space $\displaystyle \mathbb{P}$ and the subsets $\displaystyle \mathbb{V}\;of\;\mathbb{P}$ consisting of those vectors (polynomials) $\displaystyle x$ for which

    (a) x has degree 3,

    (b) 2x(0)=x(1)

    (c) $\displaystyle x(t)\geq0$ whenever $\displaystyle 0\leq t\leq 1$,

    (d) $\displaystyle x(t)=x(1-t)$ for all $\displaystyle t$.

    In which of these cases is $\displaystyle \mathbb{V}$ a vector space?

    attempt:
    I first assume that $\displaystyle \mathbb{P}$ is a vector space over $\displaystyle \mathbb{C}$ as this is not mentioned in the problem.

    (a) Not a vector space.
    $\displaystyle x(t)=a_3t^3+a_2t^2+a_1t+a_0$
    Closed under vector addition.
    Closed under scalar multiplication.
    It does not have an origin that is a polynomial of degree 3 such that $\displaystyle x(t)+0(t)=x(t)$.

    (b) Is a vector space.
    $\displaystyle x(t) = a_nt^n+\cdots+a_1t+(a_n+\cdots+a_1) $
    Closed under vector addition.
    Closed under scalar multiplication.
    There exists an element (origin) $\displaystyle 0(t)$, such that $\displaystyle x(t)+0(t)=x(t)$
    $\displaystyle 2\cdot0(0)=0(1)$

    (c) and (d), could need a push in the right direction here.

    Thanks!

    (c) Take an element here and multiply it by the scalar -1

    (d) This is a subspace, and you've to check it directly.

    Tonio
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  3. #3
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    Quote Originally Posted by Mollier View Post
    Hi,

    problem:
    Consider the vector space $\displaystyle \mathbb{P}$ and the subsets $\displaystyle \mathbb{V}\;of\;\mathbb{P}$ consisting of those vectors (polynomials) $\displaystyle x$ for which

    (a) x has degree 3,

    (b) 2x(0)=x(1)

    (c) $\displaystyle x(t)\geq0$ whenever $\displaystyle 0\leq t\leq 1$,

    (d) $\displaystyle x(t)=x(1-t)$ for all $\displaystyle t$.

    In which of these cases is $\displaystyle \mathbb{V}$ a vector space?

    attempt:
    I first assume that $\displaystyle \mathbb{P}$ is a vector space over $\displaystyle \mathbb{C}$ as this is not mentioned in the problem.

    (a) Not a vector space.
    $\displaystyle x(t)=a_3t^3+a_2t^2+a_1t+a_0$
    Closed under vector addition.
    No, it is not closed under vector addition.
    If $\displaystyle x(t)= -t^3+ 2t^2+ 3t+ 1$ and $\displaystyle y(t)= t^3+ 3t^2- t+ 2$ then $\displaystyle x(t)+ y(t)= 5t^2+ 2t+3$ which is NOT third degree.

    Closed under scalar multiplication.
    No, it is not closed under scalar multiplication. 0 is a scalar and 0 time any third degree polynomial is NOT a third degree polynomial.

    It does not have an origin that is a polynomial of degree 3 such that $\displaystyle x(t)+0(t)=x(t)$.
    But of course, that just says this is NOT a vector space- but you could have stopped after showing it is not closed under addition.

    (b) Is a vector space.
    $\displaystyle x(t) = a_nt^n+\cdots+a_1t+(a_n+\cdots+a_1) $
    Closed under vector addition.
    Closed under scalar multiplication.
    You need to show these, not just say they are true. If 2x(0)= x(1), 2y(0)= y(1) and z(t)= x(t)+ y(t), then 2z(0)= 2(x(0)+ y(0))= 2x(0)+ 2y(0)= x(1)+ y(1)= z(1) and, for any scalar, a, 2(az(t))= 2(ax(t)+ ay(t))= a(2x(t))+ a(2y(t))= ax(1)+ ay(1)= a(x(1)+ y(1))= az(1).

    There exists an element (origin) $\displaystyle 0(t)$, such that $\displaystyle x(t)+0(t)=x(t)$
    $\displaystyle 2\cdot0(0)=0(1)$
    Good! You showed that the 0 function is in this set.

    (c) and (d), could need a push in the right direction here.
    Suppose x(t) is such that $\displaystyle x(t)\ge 0$ for [/tex]0\le t\le 1[/tex]. What about (-1)x?

    (d) is pretty much like the others. If x(t)= x(1-t), y(t)= y(1-t), and z= x+ y, then z(t)= x(t)+ y(t)= x(1-t)+ y(1-t)= z(1- t), etc.
    Thanks!
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  4. #4
    Member Mollier's Avatar
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    (c)
    $\displaystyle (-1)x(t) \leq 0 $, i.e. not closed under scalar multiplication.

    (d)
    Scalar multiplication:
    $\displaystyle
    \begin{aligned}
    az(t)=ax(t)+ay(t)=&\;a(x(t)+y(t))\\
    =&\;a(x(1-t)+y(1-t))\\
    =&\;az(1-t)
    \end{aligned}
    $

    Origin:
    $\displaystyle z(t)+0(t)=z(t) $

    Now I guess I need to show that $\displaystyle 0(t)=0(t-1) $.

    $\displaystyle
    \begin{aligned}
    z(t)+0(t)=&\;z(t-1)+0(t-1)\\
    z(t)-z(t-1)+0(t)=&\;z(t-1)-z(t-1)+0(t-1)\\
    0(t)=&\;0(t-1)
    \end{aligned}
    $

    Again, thank you very much for dumbing down replies for me HallsOfIvy
    I dream of the day you won't have to!
    Last edited by Mollier; Jan 13th 2010 at 08:39 PM.
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