TITLE: Prove that these functions are linearly independent
QUESTION [6 upvotes]: Given $ \alpha_1 > \alpha_2 > \cdots > \alpha_n \geq 0$, $f_1(x)= e^{-\alpha_1x},..., f_n(x)=e^{-\alpha_nx}$, prove that they are linearly independent . Hint: don’t forget the limit $ x \rightarrow \infty $.
I have already seen proofs about similar questions. What you have to do is to set the combination of those functions (or polynomials) equal to 0 and show that all coefficients have to be 0. But here, the coefficients are in the exponent, they can’t be equal to each other, and the negative exponent disturbs me a bit.
Thanks for your help.

REPLY [2 votes]: Actually, the proof given by mechanodroid allows us to prove a more general theorem :

Eigenvectors associated with different eigenvalues of a linear operator are linearly independent.

Let's assume it for now, and note $D$ the linear function $D : f \rightarrow f'$ that returns the derivative of a function.
Now each $x\rightarrow \exp(\alpha_ix)$ is an eigenvector of $D$ (with eigenvalue $\alpha_i$) and are therefore independant.
Another consequence could be to prove that $x\rightarrow x^{\alpha_i}$ are independant as eigenvectors of $D_1:f\rightarrow(x\rightarrow x f'(x))$.

Now the proof itself consists in taking $n$ eigenvectors of $D$. Assume a linear combination of $n$ eigenvectors of $D$ is $0$. Calling $\alpha_i,f_i$ a pair eigenvalue, eigenvector, we have:
$a_1 f_1 + ... + a_n f_n = 0$
Applying $D$ once :
$\alpha_1 a_1 f_1 + ... + \alpha_n a_n f_n = 0$
Iterating, for all $i$ :
$\alpha_1^i a_1 f_1 + ... + \alpha_n^i a_n f_n = 0$
We have the following system in matrix form :
$$\left(\begin{array}{ccc}
\alpha_{1} & \ldots & \alpha_{n}\\
\vdots &  & \vdots\\
\alpha_{1}^{n} & \ldots & \alpha_{n}^{n}
\end{array}\right)\left(\begin{array}{c}
a_{1}f_{1}\\
\\
a_{n}f_{n}
\end{array}\right)=\left(\begin{array}{c}
0\\
\\
0
\end{array}\right)$$
Now the key is to prove that the matrix is invertible, so that $a_i f_i=0$ for every $i$. But its determinant is: 
$$\alpha_1 ... \alpha_n \prod_{1 \le i < j \le n} (\alpha_i - \alpha_j) $$
Which is non zero if the $\alpha$'s are different, and non zero.