1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    No, that's not even true for numbers: av+ bv= (a+b)v whether v is a vector or a number.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proving a set is a subspace
  1. Proving sets (Replies: 8)

Loading...