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Proving that g1,g2,g3 are linearly independent

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Let V = {differentiable f:R -> R}, a vector space over R. Take g1,g2,g3 in V where g1(x) = e[tex]^{}x[/tex], g2(x) = e[tex]^{}2x[/tex] and g3(x) = e[tex]^{}3x[/tex].
    Show that g1, g2 and g3 are distinct.


    2. Relevant equations
    If g1-g3 are linearly independent, it means that for any constant, k in F (field) then they all = 0 when g1k1 + g2k2 + g3k3 = 0.


    3. The attempt at a solution
    I have the idea to choose a value of x in R so that g1(x), g2(x), g3(x) are distinct, but I'm not exactly sure where to go from there because apart from choosing x=0 then for the rest of the time then they must be distinct as they all have different values. I'm just not sure how I prove this.
     
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  3. Feb 7, 2009 #2

    CompuChip

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    If they are linearly dependent, there should be choices of {k1, k2, k3} such that k1 g1(x) + k2 g2(x) + k3 g3(x) = 0 for all x (by definition of the vector (k1 g1 + k2 g2 + k3 g3).

    So probably the easiest way is to take x = 0 and x = ln(2), for example, and write down a system of equations for {k1, k2, k3} and show that only the trivial solution k1 = k2 = k3 = 0 is possible.
     
  4. Feb 7, 2009 #3

    HallsofIvy

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    Another way to do that is to use the Wronskian. If f(x), g(x), and h(x) are dependent functions, then there exist a, b, c, not all 0, so that af(x)+ bg(x)+ ch(x)= 0 for all x. One way to see if that is true is to choose 3 values for x, as CompuChip suggests, so you have three numeric equations you can solve and see if a=b=c= 0 is the only solution (independent) or if there are other solutions (dependent).

    But with functions, you can also get three equations by differentiating. If af(x)+ bg(x)+ ch(x)= 0 for all x, then it is a constant and so its derivative is also 0: af'(x)+ bg'(x)+ ch'(x)= 0. Since that is also identically 0, it is a constant and so its derivative is 0" af"(x)+ bg"(x)+ ch"(x)= 0. Taking x to be any convenient value, say 0 here, we have the three equations af(0)+ bg(0)+ ch(0)= 0, af'(0)+ bg'(0)+ ch'(0)= 0, and af"(0)+ bg"(0)+ ch"(0)= 0. That system of equations will have a unique solution, and the functions will be independent, if and only if the determinant of the array of coefficients,
    [tex]\left|\begin{array}{ccc}f(0) & g(0) & h(0) \\ f'(0) & g'(0) & h'(0) \\ f"(0) & g"(0) & h"(0)\end{array}\right|[/tex]
    the "Wronskian" of the functions at x= 0, is non-zero and the functions will be dependent if that determinant is 0.
     
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