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Homework Help: Orthogonal Basis and Inner Products

  1. Mar 3, 2012 #1
    1. The problem statement, all variables and given/known data


    3. The attempt at a solutionSince A is a vector in V and since the [itex]A_i[/itex] form a basis, we can write A as a linear combination of the [itex]A_i[/itex]. We write [itex]A = x_1 A_1 + ... + x_n A_n[/itex]. Thus, we have,

    [itex]<x_1 A_1 + ... + x_n A_n,A_i> = 0 = x_1 <A_1,A_i> + ... + x_n <A_n,A_i>[/itex]. Because two orthogonal vectors, when multiplied via inner product together give the zero vector, we simplify,

    [itex]0 = x_i <A_i,A_i>[/itex]. Because we have presumed that the [itex]A_i ≠ 0[/itex], we cannot have [itex]<A_i,A_i> = 0[/itex], so we must have that the [itex]x_i = 0[/itex]. So we have [itex]<A,A_i> = <x_1 A_1 + ... + x_n A_n,A_i> = 0 = x_1 <A_1,A_i> + ... + x_n <A_n,A_i> = x_i <A_i,A_i> = 0 <A_i,A_i> = <0,A_i> = 0. Which shows that A = 0.

    I don't know if I've done this correctly. I feel like it's not entirely convincing towards the end. Advice?
  2. jcsd
  3. Mar 3, 2012 #2


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    you're fine up to here.

    at this point, stop thinking about the "inner products", and start thinking about which linear combination of the Ai, A has to be.
  4. Mar 3, 2012 #3


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    You can say more than that- since you are told that the basis is "orthonormal", you know that [itex]<A_i, A_i>= 1[/itex]

    Once you have [itex]x_i= 0[/itex] for all x, you have immediately that [itex]A= 0A_1+ 0A_2+ ...= 0[/itex]
  5. Mar 3, 2012 #4
    Okay. That makes sense. Thank you both.

    This problem is related to the idea of the inner product. I don't think I need any help on the first parts, but I'd like to talk about the third part of the problem.


    So, finding the length of a given vector given this inner product:
    [itex]<(x,y),(x,y)> = 5x^2 + y^2[/itex].

    Taking the length, we have

    [itex]|(x,y)| = \sqrt{5x^2 + y^2}[/itex], which we define as equaling 1. Squaring both sides we find,

    [itex]5x^2 + y^2 = 1[/itex]. I think this is the equation of the circle, but I'm not sure. If it is, then my picture has y-intercepts at 1,-1 and x-intercepts at .5,-.5.

    Is this correct?
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