# How to show independent set of vectors

• Oct 1st 2008, 12:50 PM
flaming
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.
• Oct 1st 2008, 12:56 PM
Jhevon
Quote:

Originally Posted by flaming
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 $\displaystyle a = b= c = d = 0$
• Oct 1st 2008, 04:13 PM
ThePerfectHacker
Quote:

Originally Posted by flaming
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.