Results 1 to 3 of 3

Thread: another basis and dimension question

  1. #1
    Member
    Joined
    Mar 2009
    Posts
    90

    another basis and dimension question

    Let the subset $\displaystyle U \subset \mathbb{R}^4$ be defined by

    $\displaystyle U = \left \{ v | v = \left [\begin{matrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{matrix} \right ] , v_1 - v_2 + v_3 = 0 , v_2 + 2v_3 - v_4 = 0 \right \} $

    1. Show that $\displaystyle U$ is a sub-space of $\displaystyle \mathbb{R}^4$

    Because $\displaystyle \mathbb{R}^4$ is a vector space and $\displaystyle U \subset \mathbb{R}^4$ U inherits the same operations as $\displaystyle \mathbb{R}^4$. If $\displaystyle U$ is a sub-space then

    a) U is non-empty, eg. $\displaystyle \left [\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right ] \in U$

    b) $\displaystyle p \in U$ and $\displaystyle q \in U$, then $\displaystyle p + q \in U$

    $\displaystyle \left [\begin{matrix} p_4 - 3p_3 \\ p_4 - 2p_3 \\ p_3 \\ p_4 \end{matrix} \right ] + \left [\begin{matrix} q_4 - 3q_3 \\ q_4 - 2q_3 \\ q_3 \\ q_4 \end{matrix} \right ] = \left [\begin{matrix} (p_4 + q_4) - 3(p_3 + q_3) \\ (p_4 + q_4) - 2(p_3 + q_3) \\ (p_3 + q_3) \\ (p_4 + q_3) \end{matrix} \right ]$

    c) $\displaystyle \alpha \left [\begin{matrix} p_4 - 3p_3 \\ p_4 - 2p_3 \\ p_3 \\ p_4 \end{matrix} \right ] \in U$

    2. Find a basis and dimension of $\displaystyle U$

    $\displaystyle U = \left \{ v_4 \left [\begin{matrix} 1 \\ 1 \\ 0 \\ 1 \end{matrix} \right ] + v_3 \left [\begin{matrix} -1 \\ -2 \\ 1 \\ 0 \end{matrix} \right ] , v_3, v_4 \in \mathbb{R} \right \}$

    pretty simple, just the two column vectors above and so, dim(U) is 2.

    3. How do I change the basis to a basis over $\displaystyle \mathbb{R}^4$ ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1
    Hi bmp05.

    Express $\displaystyle v_3$ and $\displaystyle v_4$ in terms of $\displaystyle v_1$ and $\displaystyle v_2.$

    $\displaystyle v_3\ =\ v_2-v_1$

    $\displaystyle v_4\ =\ 2v_3+v_2\ =\ 3v_2-2v_1$

    Hence each vector in $\displaystyle U$ is of the form

    $\displaystyle \begin{bmatrix}v_1\\v_2\\v_2-v_1\\3v_2-2v_1\end{bmatrix}\ =\ v_1\begin{bmatrix}1\\0\\-1\\-2\end{bmatrix}\,+\,v_2\begin{bmatrix}0\\1\\1\\3\en d{bmatrix}$

    This shows that $\displaystyle \left\{\begin{bmatrix}1\\0\\-1\\-2\end{bmatrix},\,\begin{bmatrix}0\\1\\1\\3\end{bma trix}\right\}$ is a basis for $\displaystyle U,$ which therefore has dimension 2. (Note that it also shows that $\displaystyle U$ is a subspace of $\displaystyle \mathbb R^4.)$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,767
    Thanks
    3027
    I believe the final question was how to extend the given basis, $\displaystyle
    U = \left \{ v_4 \left [\begin{matrix} 1 \\ 1 \\ 0 \\ 1 \end{matrix} \right ] + v_3 \left [\begin{matrix} -1 \\ -2 \\ 1 \\ 0 \end{matrix} \right ] , v_3, v_4 \in \mathbb{R} \right \}
    $ to a basis for $\displaystyle R^4$. Just find two more independent vectors. That will give a set of 4 independent vectors which will form a basis for $\displaystyle R^4$. Looks to me like $\displaystyle \left[\begin{matrix}1 \\ 0 \\ 0\\ 0\end{matrix}\right]$ and $\displaystyle \left[\begin{matrix}0 \\ 1 \\ 0 \\ 0\end{matrix}\right]$ will do.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Another basis and dimension question
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Oct 3rd 2010, 06:05 PM
  2. basis and dimension
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Oct 25th 2008, 04:15 PM
  3. Basis & Dimension Question
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: Sep 20th 2008, 03:45 PM
  4. Basis and dimension question
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Jan 24th 2008, 05:25 PM
  5. Basis and Dimension
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Mar 22nd 2007, 10:03 PM

Search Tags


/mathhelpforum @mathhelpforum