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Thread: How to show independent set of vectors

  1. #1
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    How to show independent set of vectors

    Consider the vector space of all continuous functions f : [-PI; P --> R with the
    usual operations of adding two functions and multiplying a function by a scalar.
    Show that the set of vectors (functions)
    S = {sin(x); cos(x); sin(2x); cos(2x)}
    is an independent set of vectors.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by flaming View Post
    Consider the vector space of all continuous functions f : [-PI; P --> R with the
    usual operations of adding two functions and multiplying a function by a scalar.
    Show that the set of vectors (functions)
    S = {sin(x); cos(x); sin(2x); cos(2x)}
    is an independent set of vectors.
    to show that these functions are linearly independent, we must show that the only solution to $\displaystyle a \sin x + b \cos x + c \sin 2x + d \cos 2x = 0$ is $\displaystyle a = b= c = d = 0$
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  3. #3
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    Quote Originally Posted by flaming View Post
    S = {sin(x); cos(x); sin(2x); cos(2x)}
    is an independent set of vectors.
    Define an inner product on $\displaystyle C([-\pi,\pi])$ - the space of all continous functions - by $\displaystyle \left< f,g\right> = \int \limits_{-\pi}^{\pi} fg$.

    Now, $\displaystyle \{\sin x,\cos x, \sin 2x, \cos 2x\}$ is an orthogonal set of vectors.
    You need to show,
    1)$\displaystyle \int_{-\pi}^{\pi} \sin x \cos x dx = 0$
    2)$\displaystyle \int_{-\pi}^{\pi} \sin x \sin 2x dx = 0$
    3)$\displaystyle \int_{-\pi}^{\pi} \sin x \cos 2x dx = 0$
    4)$\displaystyle \int_{-\pi}^{\pi} \cos x \sin 2x dx = 0$
    5)$\displaystyle \int_{-\pi}^{\pi} \cos x \cos 2x dx = 0$
    6)$\displaystyle \int_{-\pi}^{\pi} \sin 2x \cos 2x dx = 0$

    But orthogonal sets are linearly independent. Therefore it follows this is linearly independent.

    In general we can prove that $\displaystyle \{ \bold{1}, \sin x, \cos x, ... , \sin nx, \cos nx \}$ is orthogonal (and so linearly independent) by extending the argument above.
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