Given $\sum_{i=1}^{n} \alpha_i f_i=0_E$ prove that $\alpha_1 = 0$ We have $E$ a vector space of functions $\mathbb{R} \rightarrow \mathbb{R}$
Let $a_1 > a_2 > ... > a_n$ be such that $n \geq 1$ . Let $f_1,..,f_n$ be vectors of E such that $\forall x \ \in \ \mathbb{R}, f_i(x)=\exp ^{a_ix}$. Let $\alpha_1,....,\alpha_n$ be such that $\sum_{i=1}^{n} \alpha_i f_i=0_E$
Prove that $\alpha_1=0$
I tried factorising by $e^{a_1x}$ but still do not see how to do it.
Thank you
 A: We can in fact show $\alpha_i = 0$, $1 \le i \le n$:
"Factorizing by $e^{a_1x}$" is a good way to start; that, combined with some elementary knowledge about the exponential function, will drive the solution home.
We are given that
$\sum_1^n \alpha_i e^{a_i x} = 0; \tag{1}$
we write (1) as
$e^{a_1 x}(\alpha_1 + \sum_2^n  \alpha_i e^{(a_i - a_1)x})  = 0; \tag{2}$
note we have "factored out" $e^{a_1 x}$; since $e^{a_1x}$ is never $0$,  we may divide (2) by $e^{a_1x}$ and obtain
$\alpha_1 +  \sum_2^n \alpha_i e^{(a_i - a_1)x} = 0; \tag{3}$
observing that $a_i - a_1 < 0$ for $2 \le i \le n$, we let $x \to \infty$ in (3) and see that every factor $e^{(a_i - a_1)x}$ vanishes in the limit, leaving us with
$\alpha_1 = 0. \tag{4}$
By virtue of (4), (1) now becomes
$\sum_2^n \alpha_i e^{a_i x} = 0; \tag{5}$
we may now repeat the above process and show successively that $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_n = 0$.  
The upshot is of course that the functions $e^{a_i x}$ are linearly independent over $\Bbb R$.
