Results 1 to 4 of 4

Thread: [SOLVED] Dimension

  1. #1
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318

    [SOLVED] Dimension

    $\displaystyle U_1 \cap U_2 = \{0\}$, $\displaystyle U_1 \subset V$, $\displaystyle U_2 \subset V$

    Show that $\displaystyle dim(U_1)+dim(U_2) \leq dim(V)$
    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 Spec View Post
    $\displaystyle U_1 \cap U_2 = \{0\}$, $\displaystyle U_1 \subset V$, $\displaystyle U_2 \subset V$

    Show that $\displaystyle dim(U_1)+dim(U_2) \leq dim(V)$
    Let $\displaystyle W$ be the sum of vector subspaces $\displaystyle U_1$ and $\displaystyle U_2$.
    Then $\displaystyle W\subseteq V$ and so $\displaystyle \text{dim} W \leq \text{dim} V$.
    However, $\displaystyle \text{dim} U_1 + \text{dim} U_2 = \text{dim} W$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    Quote Originally Posted by ThePerfectHacker View Post
    Let $\displaystyle W$ be the sum of vector subspaces $\displaystyle U_1$ and $\displaystyle U_2$.
    Then $\displaystyle W\subseteq V$ and so $\displaystyle \text{dim} W \leq \text{dim} V$.
    However, $\displaystyle \text{dim} U_1 + \text{dim} U_2 = \text{dim} W$.
    (1) $\displaystyle dim(U_1\oplus U_2) = dim U_1+dim U_2$
    (2) $\displaystyle U_1\oplus U_2 \subseteq V$

    $\displaystyle (1), (2) \implies \text{dim} U_1 + \text{dim} U_2 \leq \text{dim} V$

    Is that proof enough? I can prove (1), but not (2).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Spec View Post
    Is that proof enough? I can prove (1), but not (2).
    Remember what the the sum of subspaces are. It is the smallest subspace that contains both $\displaystyle U_1$ and $\displaystyle U_2$.
    However, $\displaystyle V$ already contains $\displaystyle U_1,U_2$, so $\displaystyle W\subseteq V$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. dimension
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Dec 7th 2009, 02:53 AM
  2. dimension of imT
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Jul 12th 2009, 08:58 AM
  3. dimension
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Jan 12th 2009, 05:42 AM
  4. dimension
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Apr 8th 2008, 08:01 AM
  5. PDE in three dimension
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jan 20th 2008, 08:55 PM

Search Tags


/mathhelpforum @mathhelpforum