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Subspace question

  1. Mar 6, 2012 #1
    Vector space V = Rn, S = {(x,2x,3x,...,nx) | x is a real number}

    I know that it is closed under addition, but to be closed under scalar mult.....
    say r is a scalar, then

    Say we multiply S by -4

    Then -4S = (-4x,-8x,...-4nx)

    Because the problem never stated that the number infront of x should be pos or neg, all it said was that "x is a real number" so when I mult S by a number r x is still a real number... does this explain why S is closed under scalar mult? or not exactly?
     
  2. jcsd
  3. Mar 6, 2012 #2

    Mark44

    Staff: Mentor

    S is a set of vectors. You're not multiplying S by a scalar; you're multiplying an arbitrary vector in S by a scalar.

    Let v be a vector in S, which means that v = a<1, 2, 3, ..., n> for some scalar a. Is kv also in S?
     
  4. Mar 7, 2012 #3
    well is -4v in S?

    if i multiply some vector in S by a negative scalar will that new vector still be in S?
     
  5. Mar 7, 2012 #4

    Mark44

    Staff: Mentor

    Why wouldn't it be? Every vector in S is in the form <1x, 2x, 3x, ..., nx> for some real number x. Isn't this the same as x <1, 2, 3, ..., n>?
     
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