Results 1 to 4 of 4

Math Help - Vector Spaces and Basis

  1. #1
    Junior Member
    Joined
    May 2008
    Posts
    70

    Vector Spaces and Basis

    Let S = {v_1, v_2, ..., v_n} be a finite set of vectors in a vector space V. Show that S is a basis for V iff every member of V can be written uniquely as a linear combination of the vectors in S.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    Posts
    677
    Quote Originally Posted by Coda202 View Post
    Let S = {v_1, v_2, ..., v_n} be a finite set of vectors in a vector space V. Show that S is a basis for V iff every member of V can be written uniquely as a linear combination of the vectors in S.
    This follows directly for the defintion. A set S is basis IFF
    1. It is linearly independent
    2. It spans the entire vector space

    Can you show both the above hold under the condition given in the question? It is rather straight fwd.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,993
    Thanks
    1656
    In particular, to prove the vectors are independent, saying " every member of V can be written uniquely as a linear combination of the vectors in S" means that the zero vector can be written uniquely as such a linear combination. There is one "obvious" linear combination and now you know it is the only one.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    suppose S is a basis of V.
    in case n=3
    S=\left \{ v_{1},v_{2},v_{n} \right \} is the given basis.
    let's take any x vector of V
     x\in S\Rightarrow x=\lambda_{1} v_{1}+\lambda_{2} v_{2}+\lambda_{3} v_{3}
    or x=\lambda'_{1} v_{1}+\lambda'_{2} v_{2}+\lambda'_{3} v_{3}
    We have,
    \left ( \lambda'_{1}-\lambda_{1}  \right ) v_{1}+ \left (\lambda'_{2}-\lambda _{2}  \right ) v_{2}+ \left (\lambda'_{3}-\lambda _{3}  \right ) v_{3}=0
    gives \left ( \lambda'_{1}-\lambda_{1}  \right ) =0, \left (\lambda'_{2}-\lambda _{2}  \right )=0 and \left (\lambda'_{3}-\lambda _{3}  \right ) =0 since the system \left \{ v_{1},v_{2},v_{3} \right \} is independent .
    Therefore,  \lambda'_{1}=\lambda_{1}, \lambda'_{2}=\lambda _{2} and \lambda'_{3}=\lambda _{3}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Is the intersection of two vector spaces a vector space?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 10th 2011, 11:55 AM
  2. Basis of ker L --> Basis of vector space?
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: September 17th 2011, 08:57 AM
  3. Find a basis for the vector spaces of all solutions
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 5th 2011, 05:11 AM
  4. Basis and Dimension of Vector Spaces
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 18th 2009, 06:24 PM
  5. Replies: 3
    Last Post: June 1st 2008, 01:51 PM

Search Tags


/mathhelpforum @mathhelpforum