How to show that $\{1, e^{ix}, e^{-ix}\}$ is a linearly independent set? Consider the vector space of all functions $f: \mathbb{R} \rightarrow \mathbb{C}$ over $\mathbb{C}$. If $W$ is a subspace spanned by $\beta$ = $\{1, e^{ix}, e^{-ix}\}$, show that $\beta$ is a basis for $W$.
I think I am very confused - I know I just have to show that $\beta$ is linearly independent, which means I need to show that
$a+be^{ix}+ce^{-ix} = 0 \Rightarrow a = b = c = 0$. But if I have $x=0, a=3, b=-2, c=-1$, then the left side still equals $0$ but $a, b, c \neq 0$.
I suspect I have a fundamental misunderstanding somewhere. Please advise :) 
 A: You need to show that 
$$ a + be^{ix} + ce^{-ix} = 0 \Rightarrow a, b, c = 0 $$
for all $x$.
The vector space you're considering is linear combinations of functions, so the zero on the right of the equation is the zero function, not the number zero.
A: To prove that the vectors are linearly dependent, the equality 
$$a+be^{ix}+ce^{-ix} = 0$$
should hold for some $a,b,c$ for any $x$ but you have considered a particular case with $x=0$.
Indeed we have that
$$a+be^{ix}+ce^{-ix} = 0 \quad \forall x \iff a = b = c = 0$$
and the vectors are linearly independent.
A: Suppose that there are $_1,z_2,z_3\in\mathbb C$ such that $z_1+z_2e^{ix}+z_3e^{-ix}=0$. Then:


*

*if $x=0$, we have $z_1+z_2+z_3=0$;

*if $x=\pi$, we have $z_1-z_2-z_3=0$;

*if $x=\frac\pi2$, we have $z_1+iz_2-iz_3=0$.


What does this tell you about $z_1$, $z_2$, and $z_3$?
A: The Wronskian of $\{1, e^{ix}, e^{-ix}\}$ is
$$\begin{vmatrix} 1 & e^{ix} & e^{-ix} \\ 0 & ie^{ix} & -ie^{-ix} \\ 0 & -e^{ix} & -e^{-ix}\end{vmatrix} = -2i \ne 0$$
so $\{1, e^{ix}, e^{-ix}\}$ is linearly independent.
A: While most of the answers given here are true, they don't really go into enough detail on what you understood wrong. When talking about the vectorspace of all functions, we need to consider any function $f\colon \mathbb{R}\to\mathbb{C}$ to be a vector, i.e. the given $\beta$ would be denoted much more appropriately as
$$
f_1\colon \mathbb{R}\to\mathbb{C},\,x\mapsto 1\\
f_2\colon \mathbb{R}\to\mathbb{C}, \,x\mapsto e^{ix}\\
f_3\colon \mathbb{R}\to\mathbb{C}, \,x\mapsto e^{-ix}\\
\beta=\{f_1,f_2,f_3\}$$
Now, for $f_1,f_2,f_3$ to be linearly independent, you need to show that there is no combination of $a,b,c\in\mathbb{C}$ for which the linear combination $af_1+bf_2+cf_3$ equals the $0$ in your vectorspace. Here you need to make sure though that you understand what the $0$ in your vectorspace actually is: You are looking at a vectorspace which consists of functions, i.e. your $0$ is also a function, in this case the function $n\colon \mathbb{R}\to\mathbb{C},\,x\mapsto 0$, the constant $0$-function. Now, when are 2 functions equal? When they attain the same value for every input. That's the reason you need to check for every $x\in\mathbb{R}$. The equation $af_1+bf_2+cf_3=0(=n)$ means that for every input value $x\in\mathbb{R}$ the functionvalue $af_1(x)+bf_2(x)+cf_3(x)=n(x)=0$, where $a,b$ and $c$ do not depend on $x$.
