how to show that $\{ \cos(k\omega x): 1\leq k \leq n, n\in\mathbb N \}$ is L.I? I need show that $\cos(\omega x),\cos(2\omega x), \cos(3\omega x), \ldots,\cos(n\omega x) $ is linearly independent? i'm studing linear algebra for physics and I can't use an inner product, I try to show this using the Wronskian but it seems so difficult. How can I do it? 
 A: They're eigenfunctions of $D=\frac{d^{2}}{dx^{2}}$ with different eigenvalues:
$$
             D\cos(k\omega x)=-k^{2}\omega^{2}\cos(k\omega x).
$$
Therefore, if
$$
              \sum_{k}A_k\cos(k\omega x) =0,
$$
you can show that $A_n=0$ for every $n$ because
$$
     0=\prod_{k\ne n}(D+k^{2}\omega^{2})\sum_k A_k\cos(k\omega x)=\prod_{k\ne n}(-n^{2}\omega^{2}+k^{2}\omega^{2})A_n\cos(n\omega x).
$$
A: Without loss of generality, set $\omega =1$. 
Now, since $\cos kx=\frac{1}{2}(e^{ikx}-e^{-ikx}),\ $ it suffices to prove that $\left \{ e^{ikx}-e^{-ikx} \right \}_{1\leq k\leq n}$ is an independent set. Now it's easy to compute the Wronskian.
On the other hand, we can do it directly once we make the
Observation: $\left \{ e^{ikx} \right \}_{0\leq k\leq 2n}$ is an independent set.
With this in mind, consider 
$A_{1}e^{ix}-A_{1}e^{-ix}+A_{2}e^{2ix}-A_{2}e^{-2ix}+\cdots +A_{n}e^{inx}-A_{n}e^{-inx}=0$. 
This is 
$\left ( A_{1}e^{ix}+A_{2}e^{2ix}+\cdots +A_{n}e^{nix} \right )-\left ( A_{1}e^{-ix}+A_{2}e^{-2ix}+\cdots +A_{n}e^{-nix} \right )=0$. 
Multiply both sides by $e^{nix}$ to obtain
$\left ( A_{1}e^{(n+1)ix}+A_{2}e^{(n+2)ix}+\cdots +A_{n}e^{2nix} \right )-\left ( A_{1}e^{(n-1)ix}+A_{2}e^{(n-2)ix}+\cdots +A_{n} \right )=0$.
Thus, appealing to the observation, we see that $A_1=A_2=\cdots =A_{n-1}= A_n=0$ and so our original set is independent.
