TITLE: $1, e^{ix}, e^{-ix}$ are linearly independent
QUESTION [6 upvotes]: Consider the space of all functions $f: \mathbb{R}\longrightarrow \mathbb{C}$.
Prove that $\{1, e^{ix}, e^{-ix}\}$ are linearly independent vectors.

REPLY [13 votes]: 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).