Vectors Linear Independent

  1. 1. The problem statement, all variables and given/known data
    Considering the following vectors R[tex]^{4}[/tex]:

    v1 = (1,2,0,2) v2 = (2,3,1,4) v3 = (0,1,-1,0)

    Determine if these vectors are linearly independent. Let S be the linear span of the three vectors. Define a basis and the dimensions of S. Express the vector v=(3,5,1,6) as a linear combination of the three vectors. Can this be achieved in a unique way? Justify your answer?

    2. Relevant equations
    I tried to put it into matrix form and reduce via row echolon but I'm not if this is the correct or proper way


    3. The attempt at a solution

    [ 1 2 0 2
    2 3 1 4
    0 1 -1 0
    3 5 1 6]

    [ 1 2 0 2
    0 -1 1 0
    0 1 -1 0
    0 0 0 0 ]

    x +2y = 2
    y - z = 0
    -y + 2 = 0
    therefore
    y=z making it linearly independent
     
  2. jcsd
  3. You need to prove that p=q=r=0, for v1,v2,v3 to be linear independent:

    [tex]pv_1 + qv_2 +rv_3=0[/tex]

    [tex]p(1,2,0,2)+q(2,3,1,4)+r(0,1,-1,0)=0[/tex]

    You should express the vector v in same manner as linear combination of v1,v2,v3: i.e pv1+qv2+rv3=v

    p,q,r are random scalars.

    Regards.
     
  4. so with that being said which of the two do I follow from below to work out the answer?

    a)

    1p + 2q = 0
    2p +3q +r = 0
    q - r = 0
    2p + 4q = 0

    b)

    1p + 2q = 3
    2p +3q +r = 5
    q - r = 1
    2p + 4q = 6

    and if I follow b I'm I right to think that p = 1 q =2 and r = 0
     
    Last edited: Aug 25, 2009
  5. Ok, your task have two parts,

    a) to check the linear independence of the vectors v1,v2 and v3

    b)to find out if the vector v can be represented as linear combination of the vectors v1,v2 and v3.

    So you need to solve both a) and b).

    Regards.
     
  6. a)

    1p + 2q = 0 (1)
    2p +3q +r = 0 (2)
    q - r = 0 (3)
    2p + 4q = 0 (4)

    (3) q = r
    (1) p = -2q
    put (3)and(1) into (2) 2(-2q) + 3(q) +q = -4q +3q + q = 0

    p=-2
    q = 1
    r = 1

    vectors are dependent


    b)

    1p + 2q = 3 (1)
    2p +3q +r = 5 (2)
    q - r = 1 (3)
    2p + 4q = 6 (4)

    (3) q - 1 = r
    (3) into (1) 2p + 3q + (q-1) = 5 ; 2p +4q = 6 (same as 4)
    (4) can be divide by 2 to equal (1) answer therefore is

    p = 1
    q = 1
    r = 0

    so it that then correct?

    Thank you by the way your really helpful
     
  7. I am glad that I helped you.

    Just a little correction:
    a)
    r=q
    p=-2q
    q any number in R, you chose q=1

    The vectors are linear dependent

    b)
    r=q-1
    p=3-2q
    q any number in R, you chose it q=1

    Regards.
     
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