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Linear (in)dependance

  1. Feb 1, 2009 #1
    if x1=
    (2)
    (1)
    (3)

    x2=
    (3)
    (-1)
    (4)

    x3=
    (2)
    (6)
    (4)


    show x1,x2,x3 are linear dependent and x1 and x2 are linearly dependent

    ok so i formed a 3x3 matrix
    2 3 2
    1-1 6
    3 4 4
    and got the determinant. which came to -56+42+12=0 which proves they are linearly dependent right?

    for the second part i looked for 2 constants a and b so that a(x1)+b(x2)=0
    this gives me 3 eqs. 2a+3b=0 ; a-b=0 ; 3a+4b=0 but this seems to give a=b=0? which cant be right?Cant figure out what i'm doing wrong!
     
  2. jcsd
  3. Feb 1, 2009 #2

    Mark44

    Staff: Mentor

    For the first part, your work shows that x1, x2, and x3 are linearly dependent. However, do you understand why it is that the determinant of your matrix being zero implies that the vectors are linearly dependent?

    With about the same amount of work as you did in calculating the determinant, you could have row-reduced the matrix, and from that reduced matrix you could have found that -4*x1 -2*x2 + x3 = 0. IOW, you could have shown a nontrivial solution for the equation c1*x1 + c2*x2 + c3*x3 = 0, which is how linear dependence is defined.

    For the second part, x1 and x2 are NOT linearly dependent. They are linearly independent. Any two vectors will be linearly dependent if one of them is a scalar multiple of the other. x2 is clearly not a multiple of x1 (and vice versa), so the two are linearly independent.
     
  4. Feb 1, 2009 #3
    first,thanks for the reply,appreciate it.

    if i row reduce the matrix

    i seem to get
    1 -1 6
    0 1 -2
    0 0 0
    but i'm not sure where you went from there, or do i multiply out by c1,c2,c3 first or do it now?
     
  5. Feb 1, 2009 #4
    sorry what does iow mean?
     
  6. Feb 1, 2009 #5
    should this be -4x1 -2x2 +6x3 = 0?
     
  7. Feb 1, 2009 #6

    Mark44

    Staff: Mentor

    In other words...
     
  8. Feb 1, 2009 #7

    Mark44

    Staff: Mentor

    I took it one more step and got
    1 0 4
    0 1 -2
    0 0 0

    This means that c1 = -4c3, c2 = 2c3, and c3 = c3 (i.e., c3 is arbitrary)

    If you set c3 =1, an obvious choice, you get c1 = -4, c2 = 2, and c3 = 1.

    A nontrivial solution to the equation c1*x1 + c2*x2 + c3*x3 = 0 is
    -4*x1 + 2*x2 + x3 = 0 (I omitted the sign on the first coefficient in my earlier post.)
     
  9. Feb 2, 2009 #8
    got you.

    i then did the same for the second part just to prove it i rowreduced the matrix and got

    1 -1
    0 1
    0 1

    which gives c2=0 and c1=c2=0 -vectors are linearly independent.

    then,
    what is the dimension of span (x1,x2,x3)?

    from part 1 where i have shown (x1,x2,x3) are linearly dependent does it then not follow that the dimension is 1?

    or since (x1,x2) are a linearly independent subset does this contradict this?
     
  10. Feb 3, 2009 #9
    another bit to this is-what is the dimension of span (x1,x2,x3)?

    from part 1 where i have shown (x1,x2,x3) are linearly dependent does it then not follow that the dimension is 1?

    or since (x1,x2) are a linearly independent subset does this contradict this?
     
  11. Feb 3, 2009 #10

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The dimension of the span of (x1,x2,x3) is going to be the size of a basis of this. The three vectors are linearly dependent, so you know the dimension is going to be less than 3, since every element in the span can be written as a linear combination of x1, x2 and x3, and by linear independence you can write every element in the span as a linear combination of just x1 and x2 say. x1 and x2 are linearly independent, and you should know that any linearly independent set is smaller to or has the same number of elements as the dimension of the vector space. So the dimension has to be at least 2
     
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