Results 1 to 8 of 8

Math Help - Dimensional

  1. #1
    Junior Member
    Joined
    Oct 2010
    Posts
    47

    Dimensional

    Let V be an n-dimensional vector space and let f be a nonzero linear form on V, that is a nonzero element of the dual space V* of V. Prove that the kernel of f is (n-1)-dimensional. Conversely, show that every (n-1)- dimensional linear subspace of V is the kernel of a linear form.
    Plx help...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by mathbeginner View Post
    Plx help...
    I'll give hints

    The first part is trivial if you know the rank-nullity theorem since, if V is an F-space then a linear functional \varphi:V\to F can be viewed as a linear transformation with F viewed as a one dimensional vector space over itself.

    Otherwise, try to show that if \varphi(x_0)\ne 0 then V=\text{span}\{x_0\}\oplus\ker\varphi.

    For the other part, what if W is an n-1 dimensional subspace of V and you considered some complement to it \text{span}\{x_0\}. Then, what if you considered the mapping \varphi:V\to W:w+\alpha x_0\mapsto \alpha?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2010
    Posts
    47
    Quote Originally Posted by Drexel28 View Post
    I'll give hints

    The first part is trivial if you know the rank-nullity theorem since, if V is an F-space then a linear functional \varphi:V\to F can be viewed as a linear transformation with F viewed as a one dimensional vector space over itself.

    Otherwise, try to show that if \varphi(x_0)\ne 0 then V=\text{span}\{x_0\}\oplus\ker\varphi.

    For the other part, what if W is an n-1 dimensional subspace of V and you considered some complement to it \text{span}\{x_0\}. Then, what if you considered the mapping \varphi:V\to W:w+\alpha x_0\mapsto \alpha?
    FOR the first part: as I know V*=L(V,F)
    and dimV*=dimV
    by using the rank-nullity theorem
    I get dim F(v) + dim(Ker F)
    I want ask that is it dim F(v) = dim F then I can have dim F =1???
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by mathbeginner View Post
    FOR the first part: as I know V*=L(V,F)
    and dimV*=dimV
    by using the rank-nullity theorem
    I get dim F(v) + dim(Ker F)
    I want ask that is it dim F(v) = dim F then I can have dim F =1???
    Ok, so from the above I assume you know the Rank-Nullity theorem.

    I first ask you to realize that every non-zero linear functional is surjective. The easy, non-dimension argument, comes from the fact that if \varphi(v_0)\ne0_F then for any \alpha\in F we have that \displaystyle \varphi\left(\frac{\alpha}{\varphi(v_0)}v_0\right)  =\frac{\alpha}{\varphi(v_0)}\varphi(v_0)=\alpha.

    So, you would agree that if we take the field operations of F and turn them into vector operations (i.e. the scalars and vectors are elements of F) we're left with a vector space, call it \mathcal{F}. It's fairly easy to see though that \{1_F\} is linearly independent and \text{span}\{1_F\}=\mathcal{F} and thus \dim_F \mathcal{F}=1. Note though that with this in mind we can see that any linear functional \varphi:V\to F is, in fact, also a linear transformation \varphi:V\to\mathcal{F}. But, the Rank-Nullity theorem then states that \dim_F V=\dim_F\ker\varphi+\dim_F\varphi\left(V\right). Note though that by the surjectivity of \varphi we have that \varphi\left(V\right)=\mathcal{F} and so \dim_F\varphi\left(V\right)=1 and thus plugging this into our nice little formula gives \dim_F \ker\varphi=\dim_F V-1.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2010
    Posts
    47
    Quote Originally Posted by Drexel28 View Post
    I'll give hints

    The first part is trivial if you know the rank-nullity theorem since, if V is an F-space then a linear functional \varphi:V\to F can be viewed as a linear transformation with F viewed as a one dimensional vector space over itself.

    Otherwise, try to show that if \varphi(x_0)\ne 0 then V=\text{span}\{x_0\}\oplus\ker\varphi.

    For the other part, what if W is an n-1 dimensional subspace of V and you considered some complement to it \text{span}\{x_0\}. Then, what if you considered the mapping \varphi:V\to W:w+\alpha x_0\mapsto \alpha?
    so the second part: if I say W is and n-1 dim subspace of V
    then dimV is dimW + span{x0}. But what do you mean by \varphi:V\to W:w+\alpha x_0\mapsto \alpha?[/QUOTE] what is \alpha x_0\ and why it go to \alpha\? What does that mean???
    THX
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by mathbeginner View Post
    so the second part: if I say W is and n-1 dim subspace of V
    then dimV is dimW + span{x0}. But what do you mean by \varphi:V\to W:w+\alpha x_0\mapsto \alpha?
    what is \alpha x_0\ and why it go to \alpha\? What does that mean???
    THX[/QUOTE]

    Why do I get the niggling feeling that you're trying to milk me for answers?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Oct 2010
    Posts
    47
    Quote Originally Posted by Drexel28 View Post
    what is \alpha x_0\ and why it go to \alpha\? What does that mean???
    THX
    Why do I get the niggling feeling that you're trying to milk me for answers?[/QUOTE]

    nonono..
    I am trying to understand what are u talking abt because I reli don't get what is it.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by mathbeginner View Post

    nonono..
    I am trying to understand what are u talking abt because I reli don't get what is it.
    We know that if W is a n-1 dimensional subspace that we can fix a complement U of it so that V=W\oplus U. But, since n=\dim_F W+\dim_F U we know that \dim_F U=1 and thus U=\text{span}\{x_0\} for some x_0\in V. But, since V=W\oplus U we know that every vector v\in V can be represented uniquely as w+\alpha x_0 for some w\in W and some \alpha\in F. So consider the function \varphi that takes w+\alpha x_0 and maps it to \alpha. Show that \varphi\in\text{Hom}\left(V,F\right) (if you aren't familiar with this notation, it just means V^*) and \ker\varphi=W.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. three-dimensional applications
    Posted in the Geometry Forum
    Replies: 3
    Last Post: January 11th 2011, 11:51 PM
  2. Formula of a Three-Dimensional Ray
    Posted in the Geometry Forum
    Replies: 4
    Last Post: June 26th 2010, 02:00 PM
  3. Replies: 1
    Last Post: February 25th 2010, 02:15 AM
  4. one-dimensional subspace of R^3
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 18th 2009, 01:09 PM
  5. A one dimensional...
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 22nd 2006, 12:18 AM

Search Tags


/mathhelpforum @mathhelpforum