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Linear Algebra, subset of R2 not closed under scalar multipl

  1. Aug 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Construct a subset of the x-y plane R2 that is
    (a) closed under vector addition and subtraction, but not scalar multiplication.

    Hint: Starting with u and v, add and subtract for (a). Try cu and cv

    2. Relevant equations
    vector addition, subtraction and multiplication

    3. The attempt at a solution
    i think it's impossible for a combination of two matrices of rank two to have a matrix beyond rank 2. i can get it down to rank 1 and rank 0 of course but that is still a subset of R2 and that still proves that it's closed under scalar multiplication
     
  2. jcsd
  3. Aug 25, 2016 #2
    Consider the set of all vectors S = [x, y] such that x, y are integers.

    Does that work?
     
  4. Aug 25, 2016 #3

    fresh_42

    Staff: Mentor

    How come matrices into play?
    Do you know what the vector subspaces of ##\mathbb{R}^2## are?
     
  5. Aug 25, 2016 #4
    you mean xA + yB where A and B are rank 2 matrices?
     
  6. Aug 25, 2016 #5
    well R2 is a subspace of itself right? am i wrong. can anyone clear up any of my misunderstandings? thanks
     
  7. Aug 25, 2016 #6

    fresh_42

    Staff: Mentor

    Yes. And what are the one-dimensional subspaces?
     
  8. Aug 25, 2016 #7
    it's a line. and it's still under R2
     
  9. Aug 25, 2016 #8

    fresh_42

    Staff: Mentor

    Yes. But a special kind of line.
     
  10. Aug 25, 2016 #9
    the way i understand the question is. to construct a subset of R2 that is not within R2. therefore something in R3 or above?
     
  11. Aug 25, 2016 #10
    but wouldn't that still be in R2?
     
  12. Aug 25, 2016 #11

    fresh_42

    Staff: Mentor

    Your first sentence says "construct a subset of ##\mathbb{R}^2## ..."
     
  13. Aug 25, 2016 #12
    a subset in R2 not closed under scalar multiplication... would that mean an R1 or R0? because R1 and R0 is still in R2 therefore, if the resulting matrix is still within R2, then it's closed under scalar multiplication... right?
     
  14. Aug 25, 2016 #13

    fresh_42

    Staff: Mentor

    What do you mean by a matrix in this context? A subset that is closed under addition but not under scalar multiplication cannot be a subspace.
    Therefore I asked for the subspaces, because these cannot be a solution. Any set of points ##(x,y)## is a subset.

    Edit: Let's take one single point and look where addition and subtraction gets us to.
     
  15. Aug 25, 2016 #14
    Closed under scalar multiplication means that any vector in the subset could be multiplied by a scalar and still be within the subset.
     
  16. Aug 25, 2016 #15
    i understood "closed under multiplication" correctly then. but i think what the question is asking is a subset not closed under multiplication that's why it's not making any sense to me because as fresh_42 said that in order to be a subspace, it must be closed under multiplication
     
  17. Aug 25, 2016 #16

    fresh_42

    Staff: Mentor

    There is a difference between subspace and subset. The first is a vector space itself (closed under addition and scalar multiplication) whereas a subset is just a set. But here we have to include all sums and differences, so it has to be an infinite set.

    What happens if you take a certain point into your subset? You will have to include all sums and differences as well.
    That's the smallest set you can get.
     
  18. Aug 25, 2016 #17
    thank you for the clarification. i thought subspace = subsets, forgot that subsets doesn't necessarily have the zero vector
     
  19. Aug 26, 2016 #18

    Mark44

    Staff: Mentor

    No, this makes no sense. Any subset of R2 is contained in R2. What you said here is probably due to your confusion about the difference between a subset of some space and a subspace of that space.
     
  20. Aug 26, 2016 #19
    most definitely sir!
     
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