TITLE: Linear Algebra with functions
QUESTION [15 upvotes]: 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 ?

REPLY [5 votes]: You need to show the three vectors are linearly independent. In this case I would use this trick; so that you don't need to worry about them being functions and the equality to hold for every value of $x$.
If you consider $D: \mathcal{F} \rightarrow \mathcal{F}$, the derivative operator, is an endomorphism in $\mathcal F$ (i.e. a linear map from $\mathcal{F}$ to itself). The derivatives of the three functions are
$$
Df_1=De^x=e^x=f_1
$$
$$
Df_2=De^{2x}=2e^{2x}=2f_2
$$
$$
Df_3=De^{3x}=3e^{3x}=3f_3
$$
So $f_1,f_2,f_3$ are eigenvectors of $D$, with eigenvalues  $\lambda_1=1, \lambda_2=2, \lambda_3=3$, respectively. Since $f_1, f_2, f_3$ are eigenvectors with distinct eigenvalues of the same endomorphism $D$, they are linearly independent so they form a base for $W$ and $\text{dim}W=3$.