Results 1 to 3 of 3

Math Help - Unique complement of subspace

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

    Unique complement of subspace

    Hi,

    problem:

    Can it happen that a non-trivial subspace of a vector space V (i.e., a subspace different from 0 and V) has a unique complement?


    attempt:

    My answer to this would be no in general, but I do not know if there exists some exotic subspace for which this holds.

    Here's my attempt at some sort of a proof:

    Let K and H be subspaces of V.
    Let k_1,\cdots,k_m be a basis of K.
    Let h_1,\cdots,h_n be a basis of H.

    K and H are complementary subspaces if and only if the complete set,
    k_1,\cdots,k_m,h_1,\cdots,h_n is a basis of V.

    Say now that I have K and want to find a complementary subspace. Since every set of linearly independent vectors may be used as a basis, H is not unique.

    Any suggestions are greatly appreciated.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Mollier View Post
    Hi,

    problem:

    Can it happen that a non-trivial subspace of a vector space V (i.e., a subspace different from 0 and V) has a unique complement?


    attempt:

    My answer to this would be no in general, but I do not know if there exists some exotic subspace for which this holds.

    Here's my attempt at some sort of a proof:

    Let K and H be subspaces of V.
    Let k_1,\cdots,k_m be a basis of K.
    Let h_1,\cdots,h_n be a basis of H.

    K and H are complementary subspaces if and only if the complete set,
    k_1,\cdots,k_m,h_1,\cdots,h_n is a basis of V.

    Say now that I have K and want to find a complementary subspace. Since every set of linearly independent vectors may be used as a basis, H is not unique.

    I can't understand how you may think that you've proved anything...What you now must prove is that no matter what H is (non-trivial, of course), complementing its basis to a basis of the whole space cannot be made in one unique way, and this is true ALWAYS...and here you may want to distinguish different fields: finite and infinite ones.

    Tonio


    Any suggestions are greatly appreciated.

    Thanks
    .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,454
    Thanks
    1868
    Quote Originally Posted by Mollier View Post
    Hi,

    problem:

    Can it happen that a non-trivial subspace of a vector space V (i.e., a subspace different from 0 and V) has a unique complement?


    attempt:

    My answer to this would be no in general, but I do not know if there exists some exotic subspace for which this holds
    Here's my attempt at some sort of a proof:

    Let K and H be subspaces of V.
    Let k_1,\cdots,k_m be a basis of K.
    Let h_1,\cdots,h_n be a basis of H.

    K and H are complementary subspaces if and only if the complete set,
    k_1,\cdots,k_m,h_1,\cdots,h_n is a basis of V.

    Say now that I have K and want to find a complementary subspace. Since every set of linearly independent vectors may be used as a basis, H is not unique.

    Any suggestions are greatly appreciated.

    Thanks
    No, you've shown that the basis is not unique. The problem was to show whether the subspace spanned by those other vectors is unique.

    To prove that a statement like "the complementary subspace is unique" if false you only need to prove a counter example. And it doesn't have to be very "exotic".

    Let V= R^2 and let K be {(x, y)|x= 0}, the x-axis. Let U be {(x,y)| y= x}. We can take {(1, 0)} as a basis for K and {(1 1)} as a basis for U. It's clear that those vectors are indpendent and so {(1, 0), (1, 1)} is a basis for R^2. By your definition, then, U is a "complementary" subspace to K.

    Or, take W {(x,y)|y= 2x}. Now {(1, 2)} is a basis for U and, again, {(1, 0), (1, 2)} is a basis for V. But U and W are clearly NOT the same subspace.
    Last edited by HallsofIvy; February 12th 2010 at 05:44 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. two's complement
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: January 26th 2011, 03:06 PM
  2. 10's complement
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: March 25th 2010, 07:01 PM
  3. Subspace spanned by subspace
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: February 9th 2010, 08:47 PM
  4. complement help
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: September 24th 2009, 11:49 AM
  5. 2s Complement
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: June 6th 2008, 09:00 AM

Search Tags


/mathhelpforum @mathhelpforum