Results 1 to 3 of 3

Thread: Prove that y is an element of the dual space of P.

  1. #1
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1

    Prove that y is an element of the dual space of P.

    Hi,

    problem:

    If $\displaystyle (\alpha_0,\alpha_1,\alpha_2)$ is an arbitrary sequence of complex numbers, and if $\displaystyle x$ is an element of $\displaystyle \mathbb{P}$, $\displaystyle x(t)=\sum^n_{i=0}\xi_it^i$,
    write $\displaystyle y(x)=\sum^n_{i=0}\xi_i\alpha_i$. Prove that $\displaystyle y$ is an element of $\displaystyle \mathbb{P}'$ and that ever element of $\displaystyle \mathbb{P}'$ can be obtained in this manner by a suitable choice of the $\displaystyle \alpha$'s

    $\displaystyle \mathbb{P}'$ is the collection of all linear functionals on $\displaystyle \mathbb{P}$

    attempt:

    To prove that $\displaystyle y$ is an element of $\displaystyle \mathbb{P}'$ I show that $\displaystyle y$ is a linear functional:

    $\displaystyle
    \begin{aligned}
    y(x+z)=&y\left(\sum^n_{i=0}(\xi_i+\eta_i)t^i\right )\\
    =&\sum^n_{i=0}(\xi_i+\eta_i)\alpha_i\\
    =&\sum^n_{i=0}\alpha_i\xi_i + \sum^n_{i=0}\alpha_i\eta_i\\
    =&y(x)+y(z)
    \end{aligned}
    $

    $\displaystyle
    \begin{aligned}
    y(\beta x)=&\sum^n_{i=0}\beta\xi_i\alpha_i\\
    =&\beta\sum^n_{i=0}\xi_i\alpha_i\\
    =&\beta y(x)
    \end{aligned}
    $

    Prove that every element can be obtained in this manner by a suitable choice of the $\displaystyle \alpha$'s.

    I am not sure how to prove this, but perhaps I can say that if the set of $\displaystyle \alpha$'s are a basis for $\displaystyle \mathbb{P}'$, then every element of $\displaystyle \mathbb{P}'$ can be written as $\displaystyle \sum^n_{i=0}\xi_i\alpha_i$.
    -------------------------------------------------------------------------

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,781
    Thanks
    3030
    The problem with that is that the "$\displaystyle \alpha$s" are sequences of complex numbers, not members of P' and so cannot be a basis.

    But it is easy to construct a basis:

    A basis for P itself is $\displaystyle \{1, t, t^2, \cdot\cdot\cdot, t^n\}$.

    Let $\displaystyle L_i$ be the linear functional that assigns the value 1 to $\displaystyle t^i$ and 0 to all other basis members.

    Then any member of P' can be written as $\displaystyle L= a_0L_0+ a_1L_1+ a_2L_2+ \cdot\cdot\cdot+ a_nL_n$ and $\displaystyle L(\xi_0+ \xi_1t+ \cdot\cdot\cdot+ \xi_nt^n)= a_0\xi_0+ a_1\xi_1+ \cdot\cdot\cdot+ a_n\xi_n$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1
    Helpful as always,thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Normed space and dual space
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: Jun 5th 2011, 10:46 PM
  2. Dual Space of a Vector Space Question
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Apr 16th 2011, 03:02 AM
  3. Dual space of a vector space.
    Posted in the Advanced Algebra Forum
    Replies: 15
    Last Post: Mar 6th 2011, 02:20 PM
  4. Replies: 3
    Last Post: Mar 23rd 2010, 07:05 PM
  5. vector space and its dual space
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Sep 26th 2009, 08:34 AM

Search Tags


/mathhelpforum @mathhelpforum