$1, e^{ix}, e^{-ix}$ are linearly independent Consider the space of all functions $f: \mathbb{R}\longrightarrow \mathbb{C}$.
Prove that $\{1, e^{ix}, e^{-ix}\}$ are linearly independent vectors.
 A: Recall the definition of linear independence for vector $u,v$ is
$$C_1u + C_2 v = 0\iff C_1=C_2=0.$$
Now consider 
$$C_1 \mathbf{1}+C_2 e^{ix}=0,\quad \forall x\in \mathbb{K},$$
$\mathbb{K}$ can be either $\mathbb{R}$ or $\mathbb{C}$.
The only solution is $C_1=C_2=0$.
So for your case, it would be 
$$C_1 \mathbf{1}+C_2 e^{ix} + C_3 e^{-ix}=0,\quad \forall x\in \mathbb{K}\implies C_1=C_2=C_3=0.$$
A: it's easy to see that it's equivalent to saying that $1, \sin(x), \cos(x)$ are linearly independent. (because you can get both $\sin, \cos$ as a combination.) to do this assume 
$$a + b \sin(x) + c\cos(x) = 0$$ 
as a function, so for all $x$. if so then also all its derivatives must vanish, the first one in particular, so
$$b \cos(x) - c \sin(x) = 0$$
for all $x$. but if $x = \pi$ then $\sin(\pi) = 0$ and $\cos(\pi) = -1$ hence b = 0 and so c = 0, done.
A: Assume $f(x):=a\cdot 1+b\cdot e^{ix}+c\cdot e^{-ix}=0$ for all $x$.
After multiplying with $e^{ix}$ and writing $y$ for $e^{ix}$ we obtain that 
$c+ay+by^2=0$ for all $y$ in the range of $x\mapsto e^{ix}$. This range contains at least three distinct points, and a quadratic polynomial is uniquely determined by its values at three points ...
A: Linear independence of $\{f,g,h\}$ means that $\alpha f + \beta g + \gamma h=0$ implies $\alpha=\beta=\gamma=0$. Here $f(x)=1$, $g(x)=\mathrm e^{\mathrm ix}$ and $h(x)=\mathrm e^{-\mathrm ix}$.
To see this, let's just evaluate the functions at three different points. Say $x=0$, $x=\pi/2$ and $x=\pi$. Then we get
$$\begin{align}
\alpha f(0) + \beta g(0) + \gamma h(0) &= \alpha + \beta + \gamma &= 0 &&(1)\\
\alpha f(\pi/2) + \beta g(\pi/2) + \gamma h(\pi/2) &= \alpha + \mathrm i\beta-\mathrm i\gamma &= 0 &&(2)\\
\alpha f(\pi) + \beta g(\pi) + \gamma h(\pi) &= \alpha - \beta -\gamma &= 0 &&(3)
\end{align}$$
Addint equation (1) and (3) already gives $\alpha=0$. If we insert this into (1) and (2) and divide (2) by $\mathrm i$, we get
$$\begin{align}
\beta + \gamma &= 0 &(4)\\
\beta - \gamma &= 0 &(5)
\end{align}$$
Adding (4) and (5) gives $\beta=0$, subtracting those equations gives $\gamma=0$.
Thus the three functions are linearly independent.
Note that if we had found a solution, this would not have proven linear dependence, because it doesn't exclude that at other points the functions are different (imagine if we had chosen the values $0$, $2\pi$ and $4\pi$ to evaluate the functions at).
A: I assume that you want linear independence over $\mathbb{C}$.
Hint: The three functions have different behaviors as $x\rightarrow i\infty$.
