Show linear independence of cosines We're given a subset of the vector space of continuous real functions on $\mathbb{R}$ defined as
$$
\{\cos(cx) \;\lvert\; c \in \mathbb{R},\; c > 0\}
$$
How do we show this set is linearly independent? 
I saw some other questions like this on the site, but all of them used $\cos(x)$ with no constant multiplying $x$. 
 A: Suppose that $$\forall x\in\mathbb{R}, \lambda_1\cos(c_1x)+...+\lambda_n\cos(c_nx)=0,$$ where $c_1\neq...\neq c_n$ are strictly positive real numbers. Then by differentiating two times you get
$$c_1^2\lambda_1\cos(c_1x)+...+c_n^2\lambda_n\cos(c_nx)=0.$$ Repet this process and get that $$\underset{A}{\underbrace{\begin{pmatrix}1&1&...&1\\c_1^2&c_2^2&...&c_n^2\\\vdots&\vdots&&\vdots\\c_1^{2n}&c_2^{2n}&...&c_n^{2n}\end{pmatrix}}}\begin{pmatrix}\lambda_1\cos(c_1x)\\\lambda_2\cos(c_2x)\\\vdots\\\lambda_n\cos(c_nx)\end{pmatrix}=\begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix}.$$ As $A$ is a Vandermonde matrice (so inversible), you get that $\lambda_1\cos(c_1x)=0$, etc. That implies $\lambda_i=0$ for $i\in\{1,...,n\}$.
A: Let $V$ denote the vector space of infinitely-differentiable functions on $\mathbf{R}$. Define a linear operator $T$ on $V$ by $$Tf = f''.$$
For $c > 0$, let $f_c$ be the function defined by
$$f_c(x) = \cos(cx).$$
Then
$$Tf_c = -c^2 f_c.$$
In other words, $f_c$ is an eigenvector of $T$ with eigenvalue $-c^2$. The (easy) theorem that a collection of eigenvectors corresponding to distinct eigenvalues is linearly independent now implies that $\{f_c: c >0\}$ is linearly independent.
A: Well, a set is linearly independent, any subset must be independent, so what about starting with 2 functions, that is to use c=1 and c=2 and to show that {cos(x), cos(2x)} is dependent.
Arno
