Results 1 to 6 of 6

Math Help - Basis of a Dual Space

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Basis of a Dual Space

    If I have a vector space U and a dual space U^*, with e_1,...,e_n a basis of U and e^1,....,e^n a basis of U^*, does the basis of U^* always have to have this form:

    e^i(e_j)=\delta_{ij}=\begin{cases} 1 \ \ \ \ if \ i=j\\ 0 \ \ \ \ if \ i \neq j \end{cases}

    ?

    It's easy to prove that this is a basis, but i'm wondering if the basis can only be like this.

    If it is, how can I prove it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Showcase_22 View Post
    If I have a vector space U and a dual space U^*, with e_1,...,e_n a basis of U and e^1,....,e^n a basis of U^*, does the basis of U^* always have to have this form:

    e^i(e_j)=\delta_{ij}=\begin{cases} 1 \ \ \ \ if \ i=j\\ 0 \ \ \ \ if \ i \neq j \end{cases}

    ?

    It's easy to prove that this is a basis, but i'm wondering if the basis can only be like this.

    If it is, how can I prove it?

    You can't because it isn't true: simple, for a given basis in V just choose a basis in V* which is dual to ANOTHER basis in V!

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Apr 2009
    Posts
    678
    Thanks
    1
    Quote Originally Posted by Showcase_22 View Post
    If I have a vector space U and a dual space U^*, with e_1,...,e_n a basis of U and e^1,....,e^n a basis of U^*, does the basis of U^* always have to have this form:

    e^i(e_j)=\delta_{ij}=\begin{cases} 1 \ \ \ \ if \ i=j\\ 0 \ \ \ \ if \ i \neq j \end{cases}

    ?

    It's easy to prove that this is a basis, but i'm wondering if the basis can only be like this.

    If it is, how can I prove it?
    No, I don' think so.
    Proof: We have this theorem that any set of independednt vectors can be expanded to find the basis of a finite-dim vector space. So start with a linear operator which is not in the above format. This can then be expanded to include other linear operators to form the basis of a dual space. And this bases will not be the form you suggested (as the staring operator in not in that form)

    [Edit] After seeing Tonio's post. I would rather issue a warning for my post. It might not be correct. But I think as long as e_1,...,e_n a 'fixed' basis of U (for the context of this question) I should still be right.
    Last edited by aman_cc; November 21st 2009 at 04:28 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Quote Originally Posted by tonio View Post
    You can't because it isn't true: simple, for a given basis in V just choose a basis in V* which is dual to ANOTHER basis in V!

    Tonio
    umm, what do you mean by "dual to another basis in V"?

    Also aman_cc, does a linear operator exist which is not in that form?

    T_U sends a vector to a scalar. I'll choose the scalar product and see if that gives me a basis.

    ie. e^i (e_j)=<e_j,e_j>=||e_j||^2

    This is linearly independent (since it's the only vector we have in our prospective basis) so we can extend it:

    Basis: \{ e^i, f^1,.....,f^{n-1} \} (and it doesn't matter what f^p is p=1,...,n-1 since all we want is a basis where at least one basis vector is not in the form outlined in my previous post).

    We have a basis, but we know that e^i is not in the form in my previous post. Hence we have found a basis for V* that is not in the form in my previous post.

    Is this how the argument goes (but better worded of course!)?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Apr 2009
    Posts
    678
    Thanks
    1
    Quote Originally Posted by Showcase_22 View Post
    umm, what do you mean by "dual to another basis in V"?

    Also aman_cc, does a linear operator exist which is not in that form?

    T_U sends a vector to a scalar. I'll choose the scalar product and see if that gives me a basis.

    ie. e^i (e_j)=<e_j,e_j>=||e_j||^2

    This is linearly independent (since it's the only vector we have in our prospective basis) so we can extend it:

    Basis: \{ e^i, f^1,.....,f^{n-1} \} (and it doesn't matter what f^p is p=1,...,n-1 since all we want is a basis where at least one basis vector is not in the form outlined in my previous post).

    We have a basis, but we know that e^i is not in the form in my previous post. Hence we have found a basis for V* that is not in the form in my previous post.

    Is this how the argument goes (but better worded of course!)?
    Yes - That's exactly the argument I was trying to give. But as Tonio said, given \{ e^i, f^1,.....,f^{n-1} \}, we can always find a basis of U where these linear operators operate on that basis of U as per your original form.
    @Tonio-It would be great if you can plz validate our understanding here.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by aman_cc View Post
    Yes - That's exactly the argument I was trying to give. But as Tonio said, given \{ e^i, f^1,.....,f^{n-1} \}, we can always find a basis of U where these linear operators operate on that basis of U as per your original form.
    @Tonio-It would be great if you can plz validate our understanding here.

    Well, think of it as follows: let \{v_1,...,v_n\} be any basis of V , and let \{f_1,...,f_n\} be the dual basis in V^{*} of the basis \{v_2,v_1,v_3,...,v_n\}, thus:

    f_1v_i=\left\{\begin{array}{cc}1&\,\mbox{ if }\,i=2\\0&\,\mbox{ if }\,i=1,3,...,n\end{array}\right. , f_2v_i=\left\{\begin{array}{cc}1&\,\mbox{ if }\,i=1\\0&\,\mbox{ if }\,i=2,3,...,n\end{array}\right. , ..., f_jv_i=\delta_{ij}\,,\,1\leq i\leq n\,,\,3\leq j\leq n

    Well, now forget the basis \{v_2,v_1,v_3,...,v_n\} and concentrate only on \{v_1,...,v_n\} : then, the basis  \{f_1,...,f_n\}\,\mbox{ in }\,V^{*} is NOT of the proposed form.

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Basis of dual space involving free Z-module
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 1st 2011, 01:30 PM
  2. Normed space and dual space
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: June 5th 2011, 11:46 PM
  3. Dual Space of a Vector Space Question
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 16th 2011, 04:02 AM
  4. vector space and its dual space
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: September 26th 2009, 09:34 AM
  5. Basis in dual space
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: February 15th 2008, 06:38 PM

Search Tags


/mathhelpforum @mathhelpforum