Results 1 to 5 of 5

Math Help - Linear Algebra II

  1. #1
    Member
    Joined
    Dec 2006
    Posts
    79

    Linear Algebra II

    I need some help with these proofs.

    1. Prove that the intersection of any collection of subspaces of V is a subspace of V.

    2. Suppose that U is a subspace of V. What is U + U ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by taypez View Post
    I need some help with these proofs.

    1. Prove that the intersection of any collection of subspaces of V is a subspace of V.
    The subspace must contain \bold{0} thus their intersection is non-empty.

    We need to show vector addition closure and scalar multiplication closure.
    Let \b{u},\b{v} \in W_1\cap W_2
    Then,
    \b{u}+\b{v} \in W_1 because it is a subspace since \b{u},\b{v}\in W_1.
    And,
    \b{u}+\b{v} \in W_2 because it is a subspace since \b{u},\b{v}\in W_2.
    Thus,
    \b{u}+\b{v}\in W_1\cap W_2.

    Next, let k be any scalar.
    Then,
    k\b{u}\in W_1 because it is a subspace and \b{u}\in W_1.
    And,
    k\b{u}\in W_2 because it is a subspace and \b{u}\in W_2.
    Thus,
    k\b{u}\in W_1\cap W_2.

    Quote Originally Posted by taypez View Post
    2. Suppose that U is a subspace of V. What is U + U ?
    I presume that means the direct product (or direct sum).
    Meaning,
    U+U=\{(x,y)|x,y\in U\}.
    I seems that,
    U+U will be a subspace V+V.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    If I may just say that has to be the most interesting reason for deleting a message I've ever seen...

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2006
    Posts
    79
    2) Makes sense, but this problem comes before Direct Sums. Would there be another proof perhaps using just sums?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2009
    Posts
    3
    Quote Originally Posted by ThePerfectHacker View Post
    The subspace must contain \bold{0} thus their intersection is non-empty.

    We need to show vector addition closure and scalar multiplication closure.
    Let \b{u},\b{v} \in W_1\cap W_2
    Then,
    \b{u}+\b{v} \in W_1 because it is a subspace since \b{u},\b{v}\in W_1.
    And,
    \b{u}+\b{v} \in W_2 because it is a subspace since \b{u},\b{v}\in W_2.
    Thus,
    \b{u}+\b{v}\in W_1\cap W_2.

    Next, let k be any scalar.
    Then,
    k\b{u}\in W_1 because it is a subspace and \b{u}\in W_1.
    And,
    k\b{u}\in W_2 because it is a subspace and \b{u}\in W_2.
    Thus,
    k\b{u}\in W_1\cap W_2.



    I presume that means the direct product (or direct sum).
    Meaning,
    U+U=\{(x,y)|x,y\in U\}.
    I seems that,
    U+U will be a subspace V+V.
    What would U + V be then? It is a space, but is it a subspace of V?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 1st 2011, 11:00 PM
  2. Basic Linear Algebra - Linear Transformations Help
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: December 7th 2010, 04:59 PM
  3. Replies: 2
    Last Post: December 6th 2010, 04:03 PM
  4. Replies: 7
    Last Post: August 30th 2009, 11:03 AM
  5. Replies: 3
    Last Post: June 2nd 2007, 11:08 AM

Search Tags


/mathhelpforum @mathhelpforum