Results 1 to 7 of 7

Math Help - Basis

  1. #1
    Member
    Joined
    Apr 2008
    Posts
    191

    Basis

    Hi!

    I don't understand how we can answer the following question.

    How do we: extend the linearly independent set {(1,1,1,1,1),(1,2,3,4,5)} to form a basis for R^5 ?

    Any help is appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    Find 3 more linearly independent vectors to form a basis for R^5. (You are given 2).

    Recall that it is sufficient to find n linearly independent vectors to form a basis for R^n.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2008
    Posts
    191
    Quote Originally Posted by scorpion007 View Post
    Find 3 more linearly independent vectors to form a basis for R^5. (You are given 2).

    Recall that it is sufficient to find n linearly independent vectors to form a basis for R^n.
    I'm not sure how to find thse vectors in R^5...
    Would 'any' vector do? I mean can I use {(5,4,3,2,1),(6,7,8,9,1),(2,5,8,4,6)} for instance?
    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 Roam View Post
    I'm not sure how to find thse vectors in R^5...
    Would 'any' vector do? I mean can I use {(5,4,3,2,1),(6,7,8,9,1),(2,5,8,4,6)} for instance?
    Sure, IF you can prove all the 5 together are linearly independent

    Can you?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2009
    From
    Patchogue NY (Long Island)
    Posts
    21
    Well the most simple basis to R^5
    is the following 5 vectors because they are all linearly independant.

    (1,0,0,0,0)
    (0,1,0,0,0)
    (0,0,1,0,0)
    (0,0,0,1,0)
    (0,0,0,0,1)


    You have two vectors, you need 3 more that are linearly independant.

    You can just easily use 3 of those or if you want to find more vectors you can u can just do trial and error.

    The vectors you listed are probably linear combinations of the other vectors. You can check that by doing

    v5= c1v1+c2v2+c3v3+c4v4

    Usually if the vector contains a 0 component it is independant to a vector with numbers bc the c value would have to be 0 to make the vector 0 and if that happens it is unlikely to be a linear combination of the other vectors unless the other vectors have 0 in the same place.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Apr 2008
    Posts
    191
    Okay, so can I just add the vectors {(0,0,1,0,0),(0,0,0,1,0),(0,0,0,0,1)} to the set in order to extend the linearly independent set {(1,1,1,1,1),(1,2,3,4,5)} to form a basis for R^5?

    To find out whether these vectors are linearly independent together I must test them. So, for scalars c_{1},c_{2},c_{3},c_{4},c_{5} \in R , in the linear combination:

    c_{1} (1,1,1,1,1) + c_{2}(1,2,3,4,5) + c_{3} (0,0,1,0,0) + c_{4}(0,0,0,1,0) + c_{5}(0,0,0,0,1)

    There can only be one solution to the scalars, c_{1},c_{2},c_{3},c_{4},c_{5} = 0

    I know the answer to this question isn't unique but is my answer correct? Is this a right set (and a basis for R^5)?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Jul 2006
    Posts
    364
    Yes. (You can verify it by solving the system you mentioned)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Basis of ker L --> Basis of vector space?
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: September 17th 2011, 09:57 AM
  2. Replies: 4
    Last Post: August 30th 2011, 05:48 PM
  3. Basis and co-ordinates with respect to a basis.
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 5th 2010, 08:26 AM
  4. How many different basis does this set contain?
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: January 27th 2010, 10:02 PM
  5. Basis
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: November 15th 2008, 01:08 PM

Search Tags


/mathhelpforum @mathhelpforum