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Show that a set of functions is linearly independent

  1. Nov 10, 2012 #1
    Hello everybody

    I have to show that this set of vectors a = (e-t, e-it, et, eit ) is linearly independent.

    My attempt :

    f(x) = k1 * e-t + k2 * e-it + k3 * et + k4 * eit

    f '(x) = k1 * -e-t + k2 * -ie-it + k3 * et + k4 * ieit

    f ''(x) = k1 * e-t + k2 * -e-it + k3 * et + k4 * -eit

    f(0) = k1 * 1 + k2 * 1 + k3 * 1+ k4 * 1 = 0

    f '(0) = k1 * -1 + k2 * -i + k3 * 1+ k4 * i = 0

    f ''(0) = k1 * 1 + k2 * -1 + k3 * 1+ k4 * -1 = 0

    But when I plot this in maple and reduce it I get this :

    - Image -

    They're linearly dependent but this can't be correct. So I guess I'm doing something wrong but what ?

    I would really appreciate it if someone could help me.

    I HAVE SOLVED THIS !
     

    Attached Files:

    Last edited: Nov 10, 2012
  2. jcsd
  3. Nov 10, 2012 #2

    I'm going to guess that you actually meant to write powers and that you meant dependence over the complex there...so suppose

    $$ae^t+be^{-t}+ce^{it}+de^{-it}=0\,\,\,,\,\,a,b,c,d\in\Bbb C\,\,:$$

    $$(1)\;\;t=0:\Longrightarrow a+b+c+d=0$$

    $$(2)\;\;t=2\pi:\Longrightarrow ae^{2\pi}+be^{-2\pi}+c+d=0\Longrightarrow a\left(e^{2\pi}-1\right)+b\left(e^{-2\pi}-1\right)=0$$

    $$(3)\;\;t=-2\pi:\Longrightarrow ae^{-2\pi}+be^{2\pi}+c+d=0\Longrightarrow a\left(e^{-2\pi}-1\right)+b\left(e^{2\pi}-1\right)=0$$

    From (2)-(3) you already get [itex]\,a=b=0\,[/itex] . Continue a little more to get that all the coefficients must be zero.
     
  4. Nov 12, 2012 #3
    DonAntonio's solution makes sense.

    But, OP, your solution is not clear. You should be starting with a 4x4 matrix, not a 3x4 matrix. Presumably, you would also need to write up an explanation of what those matrices have to do with the original question.
     
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