Linear Algebra with functions Basically my question is - How to check for linear independence between functions ?!
Let the group $\mathcal{F}(\mathbb{R},\mathbb{R})$
  Be a group of real valued fnctions.
i.e $\mathcal{F}(\mathbb{R},\mathbb{R})=\left\{ f:\mathbb{R}\rightarrow\mathbb{R}\right\} $
Let 3 functions $f_{1},f_{2},f_{3}$
  be given such that 
$\forall x\in\mathbb{R}\,\,\,f_{1}=e^{x},\,\,f_{2}=e^{2x},\,\,f_{3}=e^{3x}$
$W=sp(f_{1},f_{2},f_{3})$
  what is $dim(W)$
 ?
How to approach this question ? (from a linear algebra perspective)
I know that $\forall x\in\mathbb{R}\,\,\,W=\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}$
And to get the dimension I need to find the base of $W$
so I need to check whether the following holds true :
$\forall x\in\mathbb{R}\,\,\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}=0\,\Leftrightarrow\,\alpha,\beta,\gamma=0$
However when $x=0$
  I get $\alpha+\beta+\gamma=0$
  which leads to infinite amount of solutions.
How to approach this question ?

If you have
$$
              \alpha e^{x} + \beta e^{2x} + \gamma e^{3x} \equiv 0,
$$
Then you can apply the derivative operator $D$ to obtain
\begin{align}
       0 & \equiv (D-2)(D-3)\{\alpha e^{x} + \beta e^{2x} + \gamma e^{3x}\} \\
    & = (1-2)(1-3)\alpha e^{x}.
\end{align}
Therefore $\alpha=0$. Then you can apply $(D-1)(D-3)$ in order to conclude that $\beta=0$. Similarly $\gamma =0$. So $\{ e^x,e^{2x},e^{3x} \}$ is a linearly independent set of functions, which means that the dimension of $W$ is $3$.
