Whether the set of functions $(1,e^{x},e^{-x})$ linearly independent 
Are the set of functions $(1,e^{x},e^{-x})$ linearly independent?

I wrote it as an augmented matrix but it brought me to nowhere. Can somebody help me?
 A: Yes, they are independent. Fix any $a,b,c\in\mathbb{R}$ such that $$\forall x\in\mathbb{R},\qquad a+be^x + ce^{-x} = 0\ .
$$
Now:


*

*Taking $x=0$ in the above, $a+b+c=0$, or equivalently $a=-(b+c)$.

*By differentiating, you get that for all $x$, $be^x - ce^{-x}=0$. Applying to $x=0$, this gives $b=c$.


Summing up, right now we get 
$$
\forall x\in\mathbb{R},\qquad b(e^x + e^{-x}) = 2b\ .
$$
But as if $b\neq 0$ the LHS would go to $\pm\infty$ when $x\to\infty$ (while the RHS is constant), we must have $b=0$. Thus, $a=0$ and $c=0$ as well.
A: Hint: show that the equation $$a+be^x+ce^{-x}=0,\qquad \forall x \in \Bbb R$$ implies $a=b=c=0$ by evaluating at $x = 0,1,2$ for example.
A: In general, eigenvectors of distinct eigenvalues are linearly independent.
For the case at hand, the operator is differentiation. 
A: I believe this question has missing background information: it doesn't specify the vector space.
Let thus the vector space be $\mathcal{F}(\mathbb{R})$, the space of all real-valued functions. In this case, 1 is the identity function.  One way to show that $1,e^x,e^{-x}$ are linearly independent is the following.
Let $c_1,c_2,c_3\in\mathbb{R}$, and assume that
\begin{equation}
c_1 + c_2 e^{x} + c_3 e^{-x} = 0 \,\,\,\,\,(*)
\end{equation}
for all $x\in\mathbb{R}.$ Differentiate both sides of $(*)$ to get the equation
\begin{equation}
c_2 e^x + c_3^\prime e^{-x}=0,
\end{equation}
where $|c_3^\prime| = |c_3|$. If $c_2,c_3^\prime$ were to be different from zero we would get a contradiction; check, for instance, what happens when $x = 0$ and then when $x=1$. So we must have $c_2=c_3=0$ and thus $c_1=0$ as well.
