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Linear independence

  1. Apr 21, 2015 #1
    1. The problem statement, all variables and given/known data
    Given the system of vectors [itex]\cos x, \cos (x+2), \sin (x-5)[/itex]. Determine whether the system is linearly independent.

    2. Relevant equations

    3. The attempt at a solution
    If it were linearly dependent, there would exist a non-trivial linear combination, such that:
    [itex]k_1\cos x + k_2\cos (x+2) + k_3\sin (x-5) = 0[/itex]. Do not suggest the Wronski determinant, please - we haven't covered it in our Algebra, yet. There's got to be a simpler way to justify the system's (in)dependence.

    If I fix any two constants as 0, then the third one will automatically be zero, as the equality has to be true for every [itex]x\in\mathbb{R}[/itex], therefore leaving us only with a trivial solution.

    There is also a Lemma in the textbook that says whenever a sub-system turns out to be linearly dependent (provided the sub-system consists of at least two vectors, as according to another Lemma, the solution is automatically trivial if the sub-system consists of only one vector) then from there it follows the whole system is linearly dependent.

    I'm quite sure the opposite isn't true - that if any sub-system is not linearly dependent then the whole system is linearly independent.

    I'm not sure what to do here.
  2. jcsd
  3. Apr 21, 2015 #2


    Staff: Mentor

    Since the equation above is identically true for all values of x, taking the derivative of both sides will give you another equation that is also identically true. Then you'll have two equations in three unknowns (##k_1, k_2, k_3##). To get the third equation, take the derivative one more time.
  4. Apr 21, 2015 #3

    Ray Vickson

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    Use the addition formulas for ##\sin## and ##\cos##, to re-express ##\cos(x+2)## and ##\sin(x-5)## in terms of ##\cos(x)## and ##\sin(x)##.
  5. Apr 21, 2015 #4
    Ah, so I have to make sure the system determinant is not 0 and then I can safely deduce the only solution is trivial and therefore the system is linearly independent. (Attempting to solve via Cramer's method)

    Trouble: the system determinant is 0, so for every coefficient I get 0/0. According to Wronski, the system is linearly dependant if the determinant is 0, but is there another way to justifying this - we haven't gone over the Theorem in our Algebra course, yet.
    Last edited: Apr 21, 2015
  6. Apr 21, 2015 #5


    Staff: Mentor

    What I was thinking of was not Cramer's method, but writing the coefficients in a matrix and using row reduction to solve for the constants. If the system is dependent you'll get an infinite number of solutions, something that is suggested by Cramer's giving you 0 for the determinant.
  7. Apr 21, 2015 #6

    Ray Vickson

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    If you do what I suggested in #3, you will see that the three vectors u1 = cos(x), u2=cos(x+2) and u3 = sin(x-5) are linear combinations of the two vectors v1 = cos(x) and v2 = sin(x). What does that tell you about whether or not the three vectors u1, u2, u3 are linearly independent or not?
  8. Apr 21, 2015 #7
    tells me that a non-trivial solution exists and therefore the system is linearly dependant.

    Thanks, I get it now.
    Last edited: Apr 21, 2015
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