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Homework Help: Vector spaces subspaces

  1. Dec 2, 2009 #1
    1. The problem statement, all variables and given/known data

    1) Determine if
    a) (a,b,c), where b=a+c
    b) (a,b,0)
    are subspaces of R3


    2) Determine whether the given vectors span R3
    v1 = (3,1,4)
    v2 = (2,-3,5)
    v3 = (5,-2,9)
    v4 = (1,4,-1)

    2. Relevant equations

    - If u and v are vectors in W, then u + v is in W
    - If k is any scalar and u is any vector in W, then ku is in W
    - The set W of all linear combinations of v1,v2,...,vr is a subspace of V

    3. The attempt at a solution

    I know this may seems 2 relatively easy questions, i guess i'm missing something obvious here, but I was not able to figure out how to test for any subspaces and set for about 2 hours. How do we solve these kind of problems?
    Thanks, Wsaw
  2. jcsd
  3. Dec 2, 2009 #2


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    Homework Helper

    try taking arbitrary vectors and see if they're closed under the properties

    eg take
    u = (a,b,c) with a+b = c
    v = (d,e,f) with d+e = f

    what is u+v, does it satify the required property?
  4. Dec 2, 2009 #3

    but if

    u = (a,b,0)
    v = (c,d,0)


    u+v = (a+c, b+d, 0) = w
    kw = [ka+kc,kb+kd,k0)

    With that answer, I would say that all the vectors of the form (x,y,0) are subspaces of R3 but my answers book say this is not and that (x,0,0) is.
  5. Dec 2, 2009 #4


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    Homework Helper

    what does your book say?
    (x,y,0) is the z=0 plane and is a subspace
    (x,0,0) is the line along the axis and is a subspace
  6. Dec 2, 2009 #5
    My book says that all the vectors of the form (a,b,0) are not subspaces of R3 which I do not understand why.
  7. Dec 2, 2009 #6


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    Homework Helper

    doesn't sound correct to me, i think any plane through the origin shoud be a subspace, are there any other conditions?
  8. Dec 2, 2009 #7
    There are no other conditions to the vector.

    Anyway, thanks for your help!
  9. Dec 2, 2009 #8


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    Science Advisor
    Gold Member

    Properly stated one would say the set of vector of that form is a subspace. And it is provided the variables a and b are unconstrained. Double check that there were no conditions on a & b. Almost any restriction, something as simple as b not equal to 5. Will invalidate the "is a subspace" definition. (However restricting b to be a multiple of a will still yield subspace).

    Don't assume the books are always right. Assume you're wrong first but you will find errors in text books on occasion.

    One way you can approach these problems is to rewrite the variable vector as a linear combination of multiples of constant vectors. For example:

    (a,b,0) = (a,0,0) + (0,b,0) = a(1,0,0) + b(0,1,0). Thus if a and b are arbitrary you are looking at the span of the vectors (1,0,0) and (0,1,0).

    Likewise (a,b,c) with b=a+c becomes:
    (a,a+c,c) = (a,a,0) + (0,c,c) = a(1,1,0) + c(0,1,1)
    Having eliminated b you now can freely choose a and c so you are again talking about the span of a set of vectors. The span of a set of vectors in a space always defines a subspace.
    Proving that is a very good exercise BTW.
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