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Subspace of R^3

  1. Mar 24, 2009 #1
    1. The problem statement, all variables and given/known data

    Let U={(x,y,z) [tex]\in[/tex] R3 : x=z}. Show that U is a subspace of R3.



    2. Relevant equations



    3. The attempt at a solution

    U is non-empty it contains the 0 vector:

    U= {(x,y,z) = (s,t,s), s,t [tex]\in[/tex] R}

    U={s(1,0,1)+t(0,1,0), s,t [tex]\in[/tex] R}

    for s,t=0
    0(1,0,1)+0(0,1,0)=(0,0,0)

    closed under addition:

    u=(s1,t1,s1)
    v=(s2,t2,s2)

    u+v=(s1+s2,t1+t2,s1+s2)
    Hence x=z

    This was my attempt so far, I'm not sure how to prove the 3rd condition... can anyone show me how to prove that it is closed under scalar multipication?

    Regards
     
  2. jcsd
  3. Mar 24, 2009 #2
    I assume you are taking scalars in R.
    [itex]\lambda (x,y,x)=[/itex]?
     
  4. Mar 24, 2009 #3

    jambaugh

    User Avatar
    Science Advisor
    Gold Member

    You should take an arbitrary element of the set and multiply by an arbitrary scalar and then show the result is still in the set. That's what the definition means!
     
  5. Mar 25, 2009 #4
    So,

    A vector u=(s1,t1,s1) belongs to the set.

    for a scalar [tex]\lambda \in R[/tex]

    λ(s1,t1,s1)=λs1,λt1,λs1


    Is this all I need to write??
     
  6. Mar 25, 2009 #5

    Mark44

    Staff: Mentor

    Does your vector u belong to set U? If so, you should have that as your conclusion.

    Also, from your work in an earlier post, does your vector u + v belong to set U? If it does, you should say that. You concluded "hence x = z" which is not what you should conclude.
     
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