TITLE: Proof of linear independence of $e^{at}$
QUESTION [15 upvotes]: Given $\left\{ a_{i}\right\} _{i=0}^{n}\subset\mathbb{R}$ which are
distinct, show that $\left\{ e^{a_{i}t}\right\} \subset C^{0}\left(\mathbb{R},\mathbb{R}\right)$,
 form a linearly independent set of functions. 
Any tips on how to go about this proof. I tried a working from the
definition of an exponential and combining sums but that didn't seem
to get me anywhere. I saw a tip on the internet that said write it
in the form
$\mu_{1}e^{a_{1}t}+\dots+\mu_{n}e^{a_{n}t}=0$ to try to show $\mu_{1}=\dots=\mu_{n}=0$
considering each term of the left hand side must be positive, but
I can't get my head around that because while I understand $e^{x}>0\forall x\in\mathbb{R}$
I cannot see why $\mu_{i}$ must be positive in any case. I have thought
about differentiating but that doesn't seem to help. The question
did originally ask for a "rigourous" proof but I'll take any hints right now and the provided the solution of 'is obvious' is most unhelpful to me.
Any input would be fantastic. Thank you.

REPLY [7 votes]: Although the answer by Ross Millikan is probably the easiest elementary approach, the answer by Bill Dubuque points at a more profound reason that these exponential functions must be linearly independent functions: they are eigenvectors (eigenfunctions) of the differentiation operation $D:f\mapsto f'$ for distinct eigenvalues $a_1,\ldots,a_n$. It is therefore an instance of the fundamental fact that eigenspaces for distinct eigenvalues always form a direct sum. The essence of the argument can be formulated without any advanced language as follows.
We can prove the linear independence by induction of the number $n$ of distinct exponentials involved; the cases $n\leq1$ are trivial (an exponential function is not the zero function). Then by the induction hypothesis one can assume $e^{a_1x},\ldots,e^{a_{n-1}x}$ to be linearly independent. Now if $e^{a_1x},\ldots,e^{a_nx}$ were linearly dependent, the dependency relation must involve the final exponential $e^{a_nx}$ with nonzero coefficient, and therefore (after division by the coefficient) allow that function to be expressed as linear combination of $e^{a_1x},\ldots,e^{a_{n-1}x}$:
$$
  c_1e^{a_1x}+\ldots+c_{n-1}e^{a_{n-1}x}=e^{a_nx}
$$
Now (restricting to the subspace of differentiable functions, where all our exponentials obviously live), the operator $D-a_nI: f\mapsto f'-a_nf$ has the property of annihilating the final exponential function $f=e^{a_nx}$, but multiplying all other exponentials by a nonzero constant (namely $a_i-a_n$ in the case of $f=e^{a_ix}$). Moreover this operator is linear so it can be applied term-by-term; application to both sides of our identity turns it into
$$
  c_1(a_1-a_n)e^{a_1x}+\ldots+c_{n-1}(a_{n-1}-a_n)e^{a_{n-1}x}=(a_n-a_n)e^{a_nx}=0.
$$
But by the (induction) hypothesis of linear independence this can only be true if all the coefficients $c_i(a_i-a_n)$ on the left are zero, which means that all $c_i$ are zero. But in view of our original expression that is absurd. So $e^{a_1x},\ldots,e^{a_nx}$ cannot be linearly dependent, completing the induction step and the proof.