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Show that 2 set are bases for V

  1. Mar 12, 2014 #1
    1. The problem statement, all variables and given/known data


    Let V be the subspace of R3 defined by V = {(x,y,z) | x - y +2z = 0}
    Then A = {(2,0,-1) , (1,1,0)} B = {(1,3,1) , (3,1,-1)} are both bases for V.
    Show.



    3. The attempt at a solution


    1) check that both set A and set B of vectors are in R3.

    V = (x , y , (y-x)/2)

    x = 2, y = 0, z= -1 = (0-2)/2 = -1
    x = 1, y=1, z = 0 = (1-1)/2 = 0
    vectors in set A are in V

    x = 1, y = 1, z = (3-1)/2 = 1
    x = 3, y= 1, z = (1-3)/2 = 1

    vectors in set B are in V

    2) check if set A and set B are LI

    (x, y, (y-x)/2) = λ1(2, 0, -1) +λ2(1,1,0)

    Since one vector (2,0,-1) is not a multiple of another (1,1,0), it can be concluded that set A is ~LD. So LI.

    (x, y, (y-x)/2) = λ1(1,3,1) + λ2(3,1,1)

    ~LD. LI.

    3) Check that set A and set B are both spanning set for V.

    (x, y, (y-x)/2) = λ1(2,0,-1) + λ2(1,1,0)
    λ1 = (x-y)/2 , λ2 = y
    Since λ1 and λ2 are both consistent, set A is a spanning set for V.

    (x, y, (y-x)/2) = λ1(1,3,1) + λ2(3,1,1)

    λ1 = -7x-3y λ2 = (y-3x)/2

    A and B are both Basis for V
     
    Last edited: Mar 12, 2014
  2. jcsd
  3. Mar 12, 2014 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    How did you get this? It's true but you don't say why.

    Okay, or just 2- 0- 2(-1)= 0 and 1- 1+ 2(0)= 0

    typo: y= 3, not 1. Simpler is 1- 3+ 2(1)= 0

    Another typo: z= -1, not 1. Simpler is 3- 1+ 2(-1)= 0

    This is true but the you surly didn't conclude that one is not a multiple of another from
    "(x, y, (y-x)/2) = λ1(2, 0, -1) +λ2(1,1,0)"
    You seem to be confusing two different concepts of "LI".

    (x-y)/2(2,0, -1)+ y(1, 1, 0)= (x-y, 0, (y- x)/2)+ (y, y, 0)= (x, y, (x-y)/2). Yes.

    I have no idea what numbers being "consistent" means!

     
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