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Linear independance

  1. Jun 16, 2009 #1
    linear independence

    1. The problem statement, all variables and given/known data

    given that a,b and c are linearly independant vectors determine if the following vectors are linearly independant.

    a) a,0

    b) a+b, b+c, c+a

    c) a+2b+c, a-b-c, 5a+b-c

    3. The attempt at a solution

    I'm not sure how to tackle the question in this form.

    Edit:

    a) a=0a Dependant

    [itex]
    \text{Det}\left[\left(
    \begin{array}{ccc}
    1 & 0 & 1 \\
    1 & 1 & 0 \\
    0 & 1 & 1
    \end{array}
    \right)\right]=2[/itex]

    Independant

    (c)
    [itex]
    \text{Det}\left[\left(
    \begin{array}{ccc}
    1 & 1 & 5 \\
    2 & -1 & 1 \\
    1 & -1 & -1
    \end{array}
    \right)\right]=0[/itex] Dependant

    Is it ok to use those vector co-efficients in a matrix like that?
     
    Last edited: Jun 16, 2009
  2. jcsd
  3. Jun 16, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    Where'd you get you those co-efficient matrices from?
    What is the definition of vectors being linearly dependent?
     
  4. Jun 16, 2009 #3

    Mark44

    Staff: Mentor

    That actually works. For example, if you row-reduce the matrix in c), you get
    [tex]\left(\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 0\end{array}\right)[/tex]

    This says that c1 = -2c3, c2 = -3c3, and c3 = c3. If you take c3 = 1, then c1 = -2 and c2 = -3. That linear combination of the vectors given in part c results in a sum of 0, thus demonstrating that the set is linearly dependent. Note spelling of "dependent" Gregg. Similar for independent.
     
  5. Jun 16, 2009 #4
    whoops
     
  6. Jun 16, 2009 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I think it is always better to use the basic definitions than try to memorize a specific method without understanding it.

    The definition of "dependent" for a set of vectors [itex]\left{v_1, v_2, \cdot\cdot\cdot v_n}[/quote] is that there are numbers, [itex]a_1, a_2, \cdot\cdot\cdot a_n[/itex], not all 0, such that [math]a_1v_1+ a_2v_2+ \cdot\cdot\cdot a_nv_n= 0[/itex].

    For the first problem, { a, 0}, take [itex]a_0= 0[/itex], [itex]a_1= 1[/itex]: [itex]a_0a+ a_10= 0(a)+ 1(0)= 0[/itex].

    For (b), with a+b, b+c, c+a, if [itex]a_1(a+b)+ a_2(b+ c)+ a_3(c+a)= 0[/itex] then [itex](a_1+ a_3)a+ (a_1+ a_2)b+ (a_2+ a_3)c= 0[/itex]. Since a, b, and c are independent, we must have [itex]a_1+ a_3= 0[/itex], [itex]a_1+ a_2= 0[/itex], and [itex]a_2+ a_3= 0[/itex]. Obviously, [itex]a_0= a_1= a_2= 0[/itex] satisfies that but is it the only solution?
     
  7. Jun 17, 2009 #6

    Mark44

    Staff: Mentor

    I couldn't agree with HallsOfIvy more, in what he said about the importance of understanding definitions as opposed to memorizing a technique without understanding why you are doing it. To often students get tangled up in the details of calculating a determinant or row reducing a matrix without understanding what it means that the matrix determinant is zero or why the matrix should be row reduced.

    The definition of linear independence of a set of vectors is stated very simply, but there is a subtlety to it that escapes many students. The only thing that distinguishes a set of linearly independent vectors from a set that is linearly dependent is whether the equation [itex]c_1 v_1 + c_2 v_2 + c_3 v_3 + ... + c_n v_n + = 0[/itex] has only one solution (independent vectors) or an infinite number of solutions (dependent vectors).
     
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