Results 1 to 7 of 7

Math Help - linear algebra- vector-basis-subspace

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    linear algebra- vector-basis-subspace

    Hi,
    Tell whether or not the following vectors are linear independent, if they are l.i., say if they generate \mathbb{R}^3 or \mathbb{R}^4 and in the contrary case characterize implicitly the subspace generated and give a basis of this subspace.

    The vectors are (1,1,2,4),(2,-1,-5,2),(1,-1,-4,0) and (2,1,1,6).
    My attempt : I formed and reduced this matrix : \begin{bmatrix} 1&2&1&2  \\ 1&-1&-1&1 \\ 2&-5&-4&1 \\ 4&2&0&6 \end{bmatrix} to this one \begin{bmatrix} 1&0&-\frac{1}{3}&\frac{4}{3} \\ 0&1&\frac{2}{3}&\frac{1}{3} \\ 0&0&0&0 \\ 0&0&0&0  \end{bmatrix}. I concluded by saying that the vectors are linear dependent (because of the 2 rows that are 0, so 2 of the 4 vectors are linear dependent between them) and that the vectors generate \mathbb{R}^2 (because 2 rows are reduced) and that a basis of \mathbb{R}^2 is \{ (1,0,0,0),(0,1,0,0) \}.
    I have some questions : first, is what I've done correct (at least logically)?
    Second question : how can I describe implicitly \mathbb{R}^2?
    Third question : is \{ (1,0,0,0),(0,1,0,0) \} a possible basis of \mathbb{R}^{2}? What about \{ (1,0,0,0),(0,0,1,0) \}? (I guess yes).
    Thank you very much.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Oct 2008
    Posts
    64
    The first step is correct, they are linearly dependent and the dimension is 2. But this doesn't mean that the subspace is R^2. There are lots of subspace of dimension 2.

    Your work basically tells you that you don't need all four vectors to generate the subspace. Two is enough. Can you choose these two vectors?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Thank you watchmath!
    The first step is correct, they are linearly dependent and the dimension is 2. But this doesn't mean that the subspace is R^2. There are lots of subspace of dimension 2.
    Yes, you're right.
    And about
    Your work basically tells you that you don't need all four vectors to generate the subspace. Two is enough. Can you choose these two vectors?
    I did chose the vectors (1,0,0,0) and (0,1,0,0) because they are the 2 columns with only a 1 and all the other elements are 0. So that was wrong?
    Maye I had to consider an amplied matrix... but I thought it wasn't worth it since I thought I had found an easy answer. (that is, I could reduce the rows of the matrix that are not all 0s.)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2008
    Posts
    64
    The two unit vectors (that you chose) might be not generators of the subspace even they generate a subspace of dimension 2.

    You need to take 2 vectors among the 4 original vectors. Since you know that the subspace has dimension two you can take any two vectors (among 4) for which the two are linearly independent.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    If I understood well, I can take (1,1,2,4) and (2,-1,-5,2) (the 2 first vectors).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Oct 2008
    Posts
    64
    Yes in this case you can choose that one.

    Now suppose we have a matrix A where the vector columns of A are c_1,c_2,c_3,c_4 and suppose by row operation we can reduce A into
    \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&0\  end{pmatrix}.

    What are the generators of the subspace generated by c_1,c_2,c_3,c_4 ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by watchmath View Post
    Yes in this case you can choose that one.

    Now suppose we have a matrix A where the vector columns of A are c_1,c_2,c_3,c_4 and suppose by row operation we can reduce A into
    \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&0\  end{pmatrix}.

    What are the generators of the subspace generated by c_1,c_2,c_3,c_4 ?
    I'd say c_1, c_2 and c_3. I can add that I'd say it to be true even if we had the matrix \begin{pmatrix}1&0&0&89\\0&1&0&0\\0&0&1&0\\0&0&0&0  \end{pmatrix}. I hope I'm right.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Vector orthogonality, orthonormal basis and subspace
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: September 29th 2011, 08:04 AM
  2. linear algebra-dimension of subspace spanned by vectors
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 12th 2009, 08:49 AM
  3. Basis and Dependent Vector - Linear Algebra
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 23rd 2008, 10:53 AM
  4. Basis in Linear Algebra
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: December 11th 2007, 12:48 AM
  5. Linear algebra: Basis, Nullspace, Subspace
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 1st 2007, 10:33 PM

Search Tags


/mathhelpforum @mathhelpforum