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Linearly independent vectors

  1. Mar 9, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose that A, B, and C are not linearly independent. Then show how the [tex]\alpha_{i}[/tex] can be computed, up to a common factor, from the scalar products of these vectors with each other.

    Hint: Suppose that there are non-zero values of the [tex]\alpha_{i}[/tex]'s that satisfy [tex]\alpha_{1}{\mathbf{A}+{\alpha_{2}{\mathbf{B}+{\alpha_{3}{\mathbf{C}}=0.[/tex] Then, taking the dot product of both sides of this equation with A will yield a set of equations that can be solved for the [tex]\alpha_{i}[/tex]'s.
    2. Relevant equations

    [tex]\alpha_{1}{\mathbf{A}+{\alpha_{2}{\mathbf{B}+{\alpha_{3}{\mathbf{C}}=0.[/tex]


    3. The attempt at a solution

    Based off the instructions and hint I think they are asking me to solve for the [tex]\alpha[/tex]s of the following three equations:

    [tex]\alpha_{1}{\mathbf{A{\cdot}A}+{\alpha_{2}{\mathbf{B{\cdot}A}+{\alpha_{3}{\mathbf{C{\cdot}A}}=0.[/tex]

    [tex]\alpha_{1}{\mathbf{A{\cdot}B}+{\alpha_{2}{\mathbf{B{\cdot}B}+{\alpha_{3}{\mathbf{C{\cdot}B}}=0.[/tex]

    [tex]\alpha_{1}{\mathbf{A{\cdot}C}+{\alpha_{2}{\mathbf{B{\cdot}C}+{\alpha_{3}{\mathbf{C{\cdot}C}}=0.[/tex]

    Or, they could be asking me to solve for [tex]\alpha_{i}[/tex] of these 3 equations:

    [tex]\alpha_{i}({\mathbf{A{\cdot}A}+{\mathbf{B{\cdot}A}+{\mathbf{C{\cdot}A}})=0.[/tex]


    [tex]\alpha_{i}({\mathbf{A{\cdot}B}+{\mathbf{B{\cdot}B}+{\mathbf{C{\cdot}B}})=0.[/tex]


    [tex]\alpha_{i}({\mathbf{A{\cdot}C}+{\mathbf{B{\cdot}C}+{\mathbf{C{\cdot}C}})=0.[/tex]

    How do I solve for alpha? Matrices? Substitution? Elimination?
    I'm not really sure what they mean by a common factor. Insight is always appreciated.

    *Edited: Thanks for pointing that out Mark44!
     
    Last edited: Mar 10, 2010
  2. jcsd
  3. Mar 10, 2010 #2

    Mark44

    Staff: Mentor

    I think the last sentence should say "... a set of equations that can be solved for ...
    Carrying out the multiplication in the first equation (and changing the alphai to ci gives this equation
    c1|A|2 + c2A.B + c3A.C = 0

    Do the same for the other two equations. This gives you three equations in the unknowns c1, c2, and c3. This system of equations can be solved by a number of ways, including elimination or using matrices.
    With regard to "common factor," suppose that 2A + B - 3C = 0. In the case, the constants are 2, 1, and -3. It will also be true that 4A + 2B - 6C = 0. In this case the constants are 4, 2, and -6. This triple of constants has a common factor of 2. Any triple of constants of the form 2t, t, -3t will work, and t is the common factor.
     
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