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Homework Help: Proving a set is a subspace

  1. Aug 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that the set:

    S = {x [itex]\in[/itex] R[itex]^{4}[/itex]| x = [itex]\lambda[/itex](2,0,1,-1)[itex]^{T}[/itex] for some [itex]\lambda[/itex] [itex]\in[/itex] R

    is a subspace of R[itex]^{4}[/itex]


    3. The attempt at a solution

    For the subspace theorem to hold, 3 conditions must be met:

    1) The zero vector must exist
    2) Closed under addition
    3) Closed under scalar multiplication

    1) If [itex]\lambda[/itex] = 0, the vector becomes (0,0,0,0)[itex]^{T}[/itex] - therefore that's the zero vector.

    2) Closure under addition is what I'm a bit confused about.

    If we define two new vectors, u and v and two scalars [itex]\alpha[/itex] and [itex]\beta[/itex] respectively.

    u + v = ?

    3) For closure under multiplication, isn't this obviously already closed? Heck it's being multiplied by a scalar quantity already.
     
  2. jcsd
  3. Aug 22, 2011 #2

    Dick

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    Well, if u=a*(2,0,1,-1)^T and v=b*(2,0,1,-1)^T can you show u+v has the form something*(2,0,1,-1)^T? What's the something? Sure 3) is kind of obvious, but you still have to say why.
     
  4. Aug 22, 2011 #3
    The scalar would be (a + b)

    So (a + b)*(2,0,1,-1)[itex]^{T}[/itex]

    Shouldn't it be (a + b)*(4,0,2,-2)[itex]^{T}[/itex] though ? Because if we do u + v, the vectors would add as well as the scalars.
     
  5. Aug 22, 2011 #4

    Dick

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    The vectors DO NOT add as well as the scalars. a*(2,0,1,-1)^T=(2a,0,a,-a)^T. b*(2,0,1,-1)^T=(2b,0,b,-b)^T. Add them. You don't get (a+b)*(4,0,2,-2)^T, do you? This is the distributive rule with the vector part constant.
     
  6. Aug 22, 2011 #5
    Dang, just when I thought I was getting the hang of this stuff. No, it doesn't become (4,0,2,-2)[itex]^{T}[/itex]

    It becomes (2a + 2b, 0, a+b, -a - b)[itex]^{T}[/itex]

    So that would actually be (a + b)*(2,0,1,-1)[itex]^{T}[/itex] when you factor it out.

    Interesting...I have to go over the notes again.

    Thanks a lot for your help Dick.
     
  7. Aug 23, 2011 #6

    HallsofIvy

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    No, that's not even true for numbers: av+ bv= (a+b)v whether v is a vector or a number.
     
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