# Linear Independence of an infinite set .

Let $e_n=\sin nx$ ($x\in [-\pi,\pi])$and let $A=\{e_i|i\in \mathbb{N}\}$. Prove that A is a linearly independent set.

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One way to do it is to prove that $A$ forms an orthogonal set with respect to the inner product give by $$\langle f,g\rangle=\int_{-\pi}^\pi f(x)g(x)dt.$$ If $A$ forms an orthogonal set, then $A$ is a linearly independent set.

To prove that $A$ forms an orthogonal set, we need to prove that $$\langle e_n,e_m\rangle=\int_{-\pi}^\pi \sin(nx)\sin(mx)dt=0\mbox{ for }m\neq n.$$ Note that $$\int_{-\pi}^\pi \sin(nx)\sin(mx)dt=\frac{1}{2}\int_{-\pi}^\pi [\cos(m t-nt)-\cos(mt-nt)]dt=...$$ Can you prove that the last integral is zero when $m\neq n$?

Edit: If $A$ forms an orthogonal set, then we claim that $A$ is a linearly independent set. To see this, suppose we have $$\sum_{i=1}^na_ie_{n_i}(x)=0\mbox{ for all }x\in[-\pi,\pi]$$ for some $a_i$ where $1\leq i\leq n$. Taking the inner product with $e_{n_j}$ where $1\leq j\leq n$, we have $$0=\langle \sum_{i=1}^na_ie_{n_i},e_{n_j}\rangle=\sum_{i=1}^na_i\langle e_{n_i}, e_{n_j}\rangle=a_j$$ for $1\leq j\leq n$. This implies that $A$ is a linearly independent set.

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I cant understand @Paul why linear independence follows from the fact that "A forms an orthogonal set with respect to the inner product". Do you mind elaborating the reason. And by the way what is meant by orthogonal set? – Abhra Abir Kundu Feb 24 '13 at 7:54
@AbhraAbirKundu General fact: if $(f_j)$ are nonzero pairwise orthogonal vectors in an inner product space, then they are linearly independent. Proof: take a linear combination $\sum x_jf_j$. For each $i$, take the inner product of the latter with $f_i$. This leaves $x_i(f_i,f_i)=x_i\|f_i\|^2=0$. So $x_i=0$. – 1015 Feb 24 '13 at 11:28

This is my answer (which I would like you all to evaluate).

Let $z=\cos x+i\sin x$ for $x\in [-\pi,\pi]$ and $A_k=\{e_i|i\in \mathbb{N},i\le k\}$

I will show that every $A_k$ is linearly independent.

If $\displaystyle \sum_{j=1}^{k}\lambda_je_j=0$ be a linear relation. Then I will show $\lambda_j=0,\forall j\le k$.

Now we know that $\sin nx=\frac{1}{2i}(z^n-\frac{1}{z^n})$.

So the relation becomes $\displaystyle \sum_{j=1}^{k}\frac{\lambda_j}{2i}(z^j-\frac{1}{z^j})=0\dots(1)$

Now $|z|=1$ which implies All the $z$ lies on the circle of unit radius.

So $z\ne 0\Rightarrow z^n\ne 0$

Multiplying eqn. (1) with $z^k$ we get,

$\displaystyle z^k\sum_{j=1}^{k}\frac{\lambda_j}{2i}(z^j-\frac{1}{z^j})=0$

$\Rightarrow \displaystyle \sum_{j=1}^{k}\frac{\lambda_j}{2i}(z^{j+k}-z^{k-j})=0$

SO we get a polynomial in $z$ valid for all $z$ such that $|z|=1$

As a polynomial of n degree cant have more that $n$ roots and as this polynomial has infinite roots so this must be the zero polynomial $\Rightarrow \lambda_j=0 ,\forall j\le k$

Now every finite subset will be a subset of some $A_k,k\in N$ .As all $A_k$ are linearly independent it implies that every finite subset will also be linearly independent.

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Yes, it works, good job, +1. – 1015 Feb 24 '13 at 13:31
Thanks @julien for evaluating it. – Abhra Abir Kundu Feb 24 '13 at 14:39