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Two questions involving linear independence

  1. Aug 7, 2005 #1
    How do you know if this:
    | 0_-8_5|
    |3_-7_4 |
    |-1_5_-4|
    | 1_-3_2|

    a linearly independent set?


    The answer at the back of the book say that it is independent, but obvious there are free variable in this matrix , thus imply a nontrival solution for AX=0, so it must be depend.

    Let:

    v( sub 1) =

    |1 |
    |-3|
    |2 |

    v( sub 2 ) =

    |-3 |
    |9 |
    |-6 |

    v( sub 3)

    |5|
    |-7|
    |h |


    for what value of h is v( sub 3) in Span{ V( sub 2), v( sub2)}?

    for what of h is {v(s1), v(s2), v(s3) } linear independent?
     
  2. jcsd
  3. Aug 7, 2005 #2

    EnumaElish

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    There is no such value of h because v1 and v2 are already dependent.
     
  4. Aug 8, 2005 #3

    The anwer at the back say "all h".
     
  5. Aug 8, 2005 #4

    EnumaElish

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    Let's see. v1, v2, v3 are lin. indep. iff "a v1 + b v2 + c v3 = 0 implies a = b = c = 0. " See this link.

    a (1,-3,2) + b (-3,9,-6) + c (5,-7,h) = 0

    a - 3b + 5c = 0
    -3a + 9b - 7c = 0
    2a -6b + hc = 0

    a = 3b - 5c
    -3(3b - 5c) + 9b - 7c = 0
    -9b + 15c + 9b - 7c = 0
    15c - 7c = 0 ===> c = 0 ===> a = 3b
    2a - 6b = 0
    6b - 6b = 0 for all b

    So a v1 + b v2 + c v3 = 0 is true for all a, all b and c = 0, i.e. it doesn't imply a = b = c = 0 (because it works with any a and b as long as c alone is zero). Therefore the vectors are not linearly independent for any h.
     
  6. Aug 8, 2005 #5

    EnumaElish

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    Your original matrix is 4x3, so the row vectors have to be dependent. But the column vectors may be independent.
     
  7. Aug 8, 2005 #6

    Hmm... From my point of view:

    The question of weather the three vector are depend or not can be resolved by solving

    |1_-3_5_0 |
    |-3_9_-7_0
    |2_-6_h_0 |


    or

    |1_-3_ 5_0|
    |0_0_8_0_ |
    |0_0_h-10_0|

    The answer at the back stated that the three vector are dependent for "all h". I simple do not see that is the case, since if h is 10, then there is a free variable. In other word, this imply a nontrival solution for X, thus the three vector must be depend. It is not the case when h is not 10. Can you tell me what is wrong with my reasoning?
     
    Last edited: Aug 8, 2005
  8. Aug 8, 2005 #7

    EnumaElish

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    Sorry, I am a slow thinker. What do you mean by "solving
    |1_-3_5_0 |
    |-3_9_-7_0
    |2_-6_h_0 |"?

    Also how do you get
    |1_-3_ 5_0|
    |0_0_8_0_ |
    |0_0_h-10_0|
    from
    |1_-3_5_0 |
    |-3_9_-7_0
    |2_-6_h_0 |?
    We seem to arrive at the same conclusion with different methods. I am trying to understand the method you have used. Please clarify if you need me to comment on it.
     
  9. Aug 8, 2005 #8

    VietDao29

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    Okay,
    Lets prove that a = b = c = 0 is not the only solotion to [tex]a v_1 + b v_2 + c v_3 = 0[/tex]
    Matrix A:
    [tex]A : = \left[ \begin{array}{ccc} 1 & -3 & 5 \\ -3 & 9 & -7 \\ 2 & -6 & h \end{array} \right][/tex]
    You will always have:
    [tex]\det{A} = 0 \mbox{, } \forall h[/tex]
    Therefore a = b = c = 0 is not the only solution to [tex]a v_1 + b v_2 + c v_3 = 0[/tex], there fore for all h, the 3 vectors v1, v2, v3 are dependent.
    h = 10 is one case that the 3 vectors are dependent, you can try another h like h = 1, pi, ...
    Viet Dao,
     
    Last edited: Aug 8, 2005
  10. Aug 8, 2005 #9

    HallsofIvy

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    Your terminology is very confusing: you talk about a "linearly independent set" but what you show is not a set at all: it looks like either a determinant or a matrix. What set of vectors are you talking about?

    You then refer to the vector (1, -3, 2) which I presume you take from the bottom row. But then you talk about (-3, 9, -6) which I don't see any where in the "matrix" and which is clearly not independent of (1, -3, 2). (-3, 9, -6)= (-3)(1, -3, 2).

    One common way of determining whether a set of vectors is independent is to write a matrix having those vectors as columns and then row reduce. Is that what is intended? Certainly the vectors (0, 3, -1, 1), (-8, -7, 5, -3), and (5, 4,-4, 2) are independent.
     
  11. Aug 8, 2005 #10

    VietDao29

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    I am trying to prove that:
    [tex] \left\{ \begin{array}{ccc} 1a - 3b + 5c & = & 0 \\ -3a + 9b - 7c & = & 0 \\ 2a - 6b + hc & = & 0 \end{array} \right[/tex](1)
    can have other solutions apart from a = b = c = 0 (this solution is obvious).
    (1) have only one set of solution iff det(A) <> 0.
    But because det(A) = 0, for all h. Then (1) can have other solutions, and therefore the 3 vectors are dependent. Am I missing something????
    Viet Dao,
     
  12. Aug 9, 2005 #11

    HallsofIvy

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    Am I missing something? What does this have to do with the original question?
     
  13. Aug 9, 2005 #12

    VietDao29

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    Err, maybe it's true that my terminology's very limited as I am not a native English speaker. And I self-study some maths (I'm going to grade 11th this year). So it may be true that I am wrong... :cry:
    Anyway, I learn that a group of vectors {v1, v2,..., vn} is linear independent is the only solution to:
    a1v1 + a2v2 + ... + anvn = 0 is ai = 0 for i = 1 .. n
    And I try to prove that there exists a, b, c (a ^ 2 + b ^ 2 + c ^ 2 <> 0) such that:
    [tex] a \left[ \begin{array}{c} 1 \\ -3 \\ 2 \end{array} \right] + b \left[ \begin{array}{c} -3 \\ 9 \\ -6 \end{array} \right] + c \left[ \begin{array}{c} 5 \\ -7 \\ h \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right][/tex]
    -----------------
    So in order to find h that makes the 3 vectors independent, I find h such that det(A) <> 0.
    Because in the book states that:
    [tex] \left\{ \begin{array}{ccc} a_1x_1 + a_2x_2 + ... + a_nx_n & = & 0 \\ b_1x_1 + b_2x_2 + ... + b_nx_n & = & 0 \\ ... \\ h_1x_1 + h2_x_2 + ... + h_nx_n & = & 0 \end{array} \right[/tex] (1)
    Say :
    [tex]A : = \left[ \begin{array}{ccc} a_1 & ... & a_n \\ \vdots & \ddots & \vdots \\ h_1 & ... & h_n \end{array} \right][/tex]
    From the book, it states : (1) has 1 set of solution iff det(A) <> 0.
    And (1) has many sets of solution iff det(A) = 0.
    And because det(A) = 0 for all h. So I conclude that there must exist another set of solution to (1) apart from a = b = c = 0.
    And because of that the 3 vectors are dependent, no mater what h is chosen.
    By the way, the original question is : For what of h is {v(s1), v(s2), v(s3) } linear independent?
    So am I... wrong :confused:?
    Viet Dao,
     
    Last edited: Aug 9, 2005
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