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Linear algebra, is W a subspace of R^2 problem

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Determine whether W is a subspace of the vector space: W={(x,y):y=ax, a is an integer} , V=R^2

    2. Relevant equations

    3. The attempt at a solution

    Is u+v in W?
    Let u = (u,au) and v = (v,av)
    u+v = (u,au) + (v,av) = (u+v, au + av) = (u+v, a(u+v))
    If x = u+v => u + v = (x,ax)
    => closure under addition

    Is cu in W?
    cu = c(u,au) = (cu,acu)
    If x = cu => cu = (x,ax)
    => closure under scalar multiplication

    => W is a subspace of R^2

    but my book says that W is not a subspace of R^2 and I'm not sure what I'm doing wrong. Does it have something to do with a being an integer? The problem before had something similar but with 2x instead of ax where a is an integer and it was a subspace so I thought this would have the same answer. Thanks for any help in advance, this is really confusing me now.
  2. jcsd
  3. Apr 3, 2013 #2


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    I think the answer in your book is wrong and you are right. 2 is an integer.
  4. Apr 3, 2013 #3


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    You haven't told us what vector addition and scalar multiplication is defined as. Are they the usual operations in ##R^2##?

    Also, it isn't clear whether W = {(x,y): y = ax for some fixed integer a} or whether (x,y) is in W if there is any integer m such that y = mx.
  5. Apr 3, 2013 #4


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    Oh, good point. If a is AN integer then it is. If a is ANY integer, then it's not.
  6. Apr 3, 2013 #5
    I'm using the test for a subspace given as

    "If W is a nonempty subset of a vector space V, then W is a subspace of V if and only if the following closure conditions hold.

    1. If u and v are in W, then u+v is in W.
    2. If u is in W and c is any scalar, then cu is in W."

    Vector addition and scalar multiplication in R^n are defined as

    "Let u = (u1,u2,u3,...,un) and v = (v1,v2,v3,...,vn) be vectors in R^n and let c be a real number. Then the sum of u and v is defined as the vector

    u+v = (u1 + v1, u2 + v2, u3 + v3,..., un+vn)

    and the scalar multiple of u by c is defined as the vector

    cu = (cu1, cu2, cu3,...,cun)."

    As for if a is a fixed integer or any integer, that is all the problem states, I'm assuming it means any integer not a fixed one though.
  7. Apr 3, 2013 #6
    That might be what I'm missing but what would be an example of an integer which would violate one of the closure axioms? All it really says is "a is an integer".
  8. Apr 3, 2013 #7


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    What happens if you add (x,mx) and (y,ny) if m and n are different integers?
  9. Apr 3, 2013 #8


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    Then they didn't really quantify the question clearly. If a is ANY integer, can you prove it's not a subspace? Do you see the difference here? If a is a single integer then the graph is a line. If a is any integer then it's a whole bunch of lines.
  10. Apr 3, 2013 #9
    Sure that definitely makes sense, I'm still not sure what my book actually means because there aren't any other questions that include that phrase to compare it to, which is strange because these are supposed to be review questions over the chapter. Would make sense of why the answer is like that though. Thanks to both of you for the quick replies!
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