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Help with vector spaces axioms

  1. Feb 19, 2013 #1
    1. The problem statement, all variables and given/known data
    for the 2x2 matrix [a 12;12 b] is it a vector space


    2. Relevant equations
    1. If u and v are objects in V, then u+v is in V
    2. u+v = v + u
    3. u+(v+w) = (u+v)+w
    4. There is an object 0 in V, called a zero vector for V, such that 0+u = u+0 = u for all u in V
    5. For each u in V, there is an object -u in V, called a negative of u, such that u+(-u) = (-u)+u = 0
    6. If k is any scalar and u is any object in V, then ku is in V
    7. k(u+v) = ku+kv
    8. (k+m)u = ku+mu
    9. k(mu) = (km)(u)
    10. 1u = u


    3. The attempt at a solution
    axioms 1,4,5,6 fail to hold.

    axiom 1:
    for any vectors u = [u1,12; 12, u2], v = [v1,12 ; 12 v2]in the space, u + v must also be in the space
    this is obviously false, since
    u + v = [u1+v1,12 + 12; 12+12, u2 + v2]
    = [u1+v1,24; 24, u2 + v2]
    which is not in the space

    axiom 4:
    let u = [u1,12; 12, u2], we need to find a 0 vector such that
    u + 0 = 0 + u = u
    but 0 must have the form [z1,12; 12, z2] to be in the space
    and so u + 0 = [u1,12; 12, u2] + [z1,12; 12, z2]
    = [u1+z1,12+12 ; 12+12, u2+z2]
    = [u1+z1, 24 ; 24, u2+z2]
    which is not equal to vector u, regardless of z1 and z2 values.

    axiom 5:
    can't have an additive inverse, if there's no 0 vector

    axiom 6:
    let u = [u1,12; 12, u2]
    according to this axiom, for any scalar k, k*u is also in the space
    but this is false for any k =/= 1:
    k*u = k*[u1,12; 12, u2]
    =[k*u1,k*12; k*12, k*u2]
    which is not in space unless k = 1

    however I am not sure if I did this correctly and need confirmation.
     
    Last edited: Feb 19, 2013
  2. jcsd
  3. Feb 19, 2013 #2
    You did them all correctly. Note that you only need to find one property that doesn't work to conclude that V isn't a vector space. So you could have just done the first one and be done with it.
     
  4. Feb 19, 2013 #3
    thanks for the confirmation :)
     
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