Results 1 to 8 of 8

Math Help - Basis for subspace from a set of vectors.

  1. #1
    Newbie os008's Avatar
    Joined
    Jun 2010
    From
    Egypt
    Posts
    4

    Basis for subspace from a set of vectors.

    Hello everyone,
    I'm studying for my exam in a few days, i stumbled on a question. I researched really hard in the reference book i use (Anton's Elementary Linear Algebra), and searched on the internet on general Basis and Vector Spaces. (I've been researching for 3 hours)

    The problem says ...

    Find a basis for the subspace of R^4 spanned by the given vectors:
    a) (1,1,-4,-3), (2,0,2,-2), (2,-1,3,2)
    b) (1,1,0,0), (0,0,1,1), (-2,0,2,2), (0,-3,0,3)

    I have a solution given by the doctor, but i just don't get it really. So i opted to research myself.

    I know how to prove if a set of vectors are a basis for a vector space, and how to get a basis for the solution of a system of equations, which is what i thought about doing here.

    For example, i thought about arranging those vectors in a matrix, row after row, then multiplying it by a 1x4 matrix of any arbitrary variables, kind of like getting the solution for a system of equations, but you're given the variables in that case. Then equate it by another arbitrary vector? Then get the solution in Linear Combination form, and the resulting vectors should be the basis, right?

    The doctor arranged that said matrix, reduced it, took the resulting rows as the basis, does that even make any sense?

    Anyway, i proceeded with my idea, it made the solution very complex to simplify because of the lone matrix i added. Just for the fun of it, i removed it, and put a zero matrix, the result was a basis of a 1-dimensional space ...

    a
    b
    c
    d
    vector

    Any help would be appreciated please, cause i feel i'm not solving methodically at all. Thank you.
    Last edited by os008; June 17th 2010 at 02:59 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Well, you're not told to find an orthonormal basis, or I'd say use the Gram-Schmidt procedure. You're just told to find a basis. What I would do is take the vectors you're given in part a), for example, set an arbitrary linear combination of them equal to zero, and solve for the coefficients. There may be more to it than that, but that's where I'd start. Your goal is to eliminate any superfluous vectors. It may be that the space spanned by the vectors in part a) is only 2-dimensional. It's certainly at least 1-dimensional!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2009
    Posts
    130
    First you can check that:
    (1,1,-4,-3), (2,0,2,-2), (2,-1,3,2)
    are linear independent.

    The subspace is {u(1,1,-4,-3)+v(2,0,2,-2)+w(2,-1,3,2)} or all vectors {(u+2v+2w,u-w,-4u+2v+3w,-3u-2v+2w)}

    You say: "find basis for the subspace which is spanned from the vector set above".

    Because the vectors are linear independent and they also span the subspace it follows that the basis of the subspace is {(1,1,-4,-3), (2,0,2,-2), (2,-1,3,2)}.

    I guess there is some mistake in the task. Do you want an basis of R^4 which is spanned by {(1,1,-4,-3), (2,0,2,-2), (2,-1,3,2)} ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie os008's Avatar
    Joined
    Jun 2010
    From
    Egypt
    Posts
    4
    Quote Originally Posted by p0oint View Post
    First you can check that:
    (1,1,-4,-3), (2,0,2,-2), (2,-1,3,2)
    are linear independent.

    The subspace is {u(1,1,-4,-3)+v(2,0,2,-2)+w(2,-1,3,2)} or all vectors {(u+2v+2w,u-w,-4u+2v+3w,-3u-2v+2w)}

    You say: "find basis for the subspace which is spanned from the vector set above".

    Because the vectors are linear independent and they also span the subspace it follows that the basis of the subspace is {(1,1,-4,-3), (2,0,2,-2), (2,-1,3,2)}.
    Indeed, i checked the independence and it passed. That makes them a basis for a 3-d sub-space? That solution makes sense. Thanks.

    I guess there is some mistake in the task. Do you want an basis of R^4 which is spanned by {(1,1,-4,-3), (2,0,2,-2), (2,-1,3,2)} ?
    I think that won't work, right? Because the number of vectors are less than the dimension of R4?

    Quote Originally Posted by Ackbeet View Post
    What I would do is take the vectors you're given in part a), for example, set an arbitrary linear combination of them equal to zero, and solve for the coefficients. There may be more to it than that, but that's where I'd start.
    I tried to do that for the second part of the question; b). Gave me the same result; independent. So they're a basis for a 4-d sub-space of R4?


    Another question please. What if any of the 2 sets are dependent? How can i find a basis please? Should i make a linear combination, and solve like you suggested Ackbeet? The resulting co-efficients would be the basis?

    Thanks again guys, really appreciate the help.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    For part a), I can guarantee you that the dimension is less than or equal to 3. If you've shown linear independence, then the dimension is exactly 3.
    For part b), I can guarantee you that the dimension is less than or equal to 4. If you've shown linear independence, then the dimension is exactly 4.

    In either case, if you've shown linear independence, then the set of vectors you started with is the basis, and you're done with the problem.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie os008's Avatar
    Joined
    Jun 2010
    From
    Egypt
    Posts
    4
    I stopped in studying yesterday just before Row and Column Spaces. Reading on today, i found this in the reference book ...

    Except for a variation in notation, the space spanned by these vectors is the row space of the matrix ...

    Reducing this matrix to row-echelon form, we obtain ...

    The nonzero row vectors in this matrix are ...

    These vectors form a basis for the row space and consequently form a basis for the subspace of R spanned by those vectors.
    I couldn't paste the matrix, and vectors because they're in a non-standard form.

    That explains the solution of the professor. Which i haven't attended any lectures for, because i had to work on those days, i'm depending on the reference book and the lectures i photocopied.

    Sorry guys, it's my fault, shame this is my first thread on the forums. Despite all the research i did yesterday, it seems the solution was just one section after where i stopped.

    I'll try to research harder next time. Thanks again.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,977
    Thanks
    1643
    a) (1,1,-4,-3), (2,0,2,-2), (2,-1,3,2)
    b) (1,1,0,0), (0,0,1,1), (-2,0,2,2), (0,-3,0,3)

    As you have already been told, if those vectors are independent, then the vectors given are a basis and no further work is needed.

    To see if they are independent, use the definition: for (a), do there exist numbers a, b, and c, not all 0, such that a(1, 1, -4, -3)+ b(2, 0, 2, -2)+ c(2, -1, 3, 2)= (0, 0, 0)? That is the same as the three equations a+ 2b+ 2c=0, a- c= 0, -4a+ 2b+ 3c= 0, and -3a- 2b+ 2c= 0. From the second equation, c= a so the other equations become 3a+ 2b=0, -a+ 2b= 0, and -a- 2b= 0. The last two equations are the same but adding that to the first equation, 2a= 0 so a= c= 0, and so 2b= 0 gives b= 0. Those three vectors are independent so the vectors given are a basis.

    For the (b), the question is "do there exist numbers a, b, c, and d, not all 0, such that a(1,1,0,0)+ b(0,0,1,1)+ c(-2,0,2,2)+ d(0,-3,0,3)= (0, 0, 0, 0)? That gives the four equations a- 2c= 0, a- 3d= 0, b+ 2c= 0, and b+ 2c+ 3d= 0. From the first two equations, c= a/2 and d= a/3 so the last two equations are b+ a= 0 and b+ 2a= 0. Subtracting the first from the second, a= 0 which immediately gives b= c= d= 0 also. Again, those four vectors are independent and so form a basis themselves. In fact, here, since there are four vectors and this is in R^4, the "subspace" spanned by those vectors is R^4 and (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), and (0, 0, 0, 1) also form a basis.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie os008's Avatar
    Joined
    Jun 2010
    From
    Egypt
    Posts
    4
    Quote Originally Posted by HallsofIvy View Post
    As you have already been told, if those vectors are independent, then the vectors given are a basis and no further work is needed.
    .
    .
    .
    Thank you for the detailed solution .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: April 7th 2011, 05:09 AM
  2. Basis perpendicular to a subspace spanned by 2 vectors
    Posted in the Advanced Algebra Forum
    Replies: 17
    Last Post: April 19th 2010, 06:06 PM
  3. Subspace, Basis
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: February 12th 2010, 04:31 AM
  4. Basis for Subspace
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: July 8th 2008, 07:44 AM
  5. Basis for a subspace
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 2nd 2008, 03:08 AM

Search Tags


/mathhelpforum @mathhelpforum