Results 1 to 4 of 4

Math Help - Spanning sets for R^3

  1. #1
    Member
    Joined
    Jun 2006
    Posts
    93

    Spanning sets for R^3

    Need to show that {(1,0,0)^T, (0,1,1)^T, (1,0,1)^T, (1,2,3)^T} is a spanning set for R^3.

    So I use alpha, beta, gamma, and another variable to show that this entire statement can be put into a linear combination. Would I just set the last variable to be equal to 0. It seems otherwise it would be a spanning set of R^4 instead of R^3.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by pakman View Post
    Need to show that {(1,0,0)^T, (0,1,1)^T, (1,0,1)^T, (1,2,3)^T} is a spanning set for R^3.
    Consider \textbf{u} =\begin{pmatrix} x\\ y\\ z \end{pmatrix}\in \mathbb{R}^3.We will show this general vector can be obtained by a linear combination of the first 3 elements. To do that we have to obtain the scalars of the combination.

    Quote Originally Posted by pakman View Post
    So I use alpha, beta, gamma, and another variable to show that this entire statement can be put into a linear combination.
    \textbf{u} = a\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} + b\begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}+c\begin{pmatrix} 1\\ 0\\ 1 \end{pmatrix}

    x = a+b
    y = b
    z = b+c
    This means (a,b,c) = (x-y,y,z-y). So we have the scalars for the linear combination.

    Quote Originally Posted by pakman View Post
    Would I just set the last variable to be equal to 0.
    No you cant directly do that. You should prove the first three elements are linearly independent. And then claim this set is adequate using the fact that the dimension of the space is 3.

    Quote Originally Posted by pakman View Post
    It seems otherwise it would be a spanning set of R^4 instead of R^3
    If a spanning set has 4 vectors, the dimension of the basis it spans need not be 4. I can put 100 more vectors in that spanning set. But that set shall still span \mathbb{R}^3. Dont forget that whether it span \mathbb{R}^3 or \mathbb{R}^4 is decided by the maximum number of linearly independent vectors in the set.

    P.S: Also remember that \mathbb{R}^3 has 3-tuples and \mathbb{R}^4 has 4-tuples
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2006
    Posts
    93
    Thanks, that helped a lot. I'm confused about one thing you said though, what is a 3-tuple?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by pakman View Post
    Thanks, that helped a lot. I'm confused about one thing you said though, what is a 3-tuple?
    Numbers like (1,2,3) or (e,233344,\pi). Basically vectors with 3 co-ordinates
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linearility/spanning sets.
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: December 10th 2010, 03:25 AM
  2. Spanning Sets
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: October 28th 2010, 04:31 PM
  3. spanning sets
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 10th 2010, 07:46 PM
  4. Spanning sets?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 15th 2009, 07:44 PM
  5. spanning sets
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 14th 2008, 04:14 AM

Search Tags


/mathhelpforum @mathhelpforum