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 ?
 A: Hint:
let $e^x=y$, $e^{2x}=y^2$, $e^{3x}=y^3$ you have:
$\alpha y +\beta y^2+ \gamma y^3=0$
where the $0$ at RHS is the zero polynomial. 
Now: when a polynomial is the zero polynomial?
In general: 

The $0$ at RHS is the neutral element for the sum of functions in the vector space, not  simply the number $0$ and this means that it is the function $f(x)=0\quad  \forall x \in \mathbb{R}$.

A: You have to prove
$$
\forall x\in\mathbb{R}:\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}=0\Leftrightarrow\alpha,\beta,\gamma=0,
$$
but I think the quantifier applies only to the part on the left side
of the $\Leftrightarrow$, like this:
$$
\left(\forall x\in\mathbb{R}:\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}=0\right)
\Leftrightarrow\,\alpha,\beta,\gamma=0.
$$
So for example $\alpha = -1, \beta = 1, \gamma = 0$ satisfies
$\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}=0$ when $x=0$,
but it doesn't satisfy the equation for all values of $x$.
If you had to prove
$$
\forall x\in\mathbb{R}:\left(\alpha e^{x}+\beta e^{2x}+\gamma e^{3x}=0
\Leftrightarrow\,\alpha,\beta,\gamma=0\right)
$$
then you would be in trouble, because that statement is not true;
but that's not how we prove independence of the functions,
so you don't need to worry about that.
A: Hint: Use Wronskian and show that the Wronskian-Determinant does not vansish.
A: 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 $\text{dim}W=3$.
A: 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 $3$.
A: You need to check if the functions are independent, as you said. 
A way to go about this, which that ties it in with things you likely know is to evaluate it at several points, as you did for $x=0$. 
You get one condition for $x=0$. You get another condition for $x=1$ and still another one for $x=2$. 
Each will allow more than one solution, but they'll only have one common solution, which is what you are after.
A: Write
$$\alpha e^x + \beta e^{2x} + \gamma e^{3x} = 0$$
You can go ahead and cancel out a positive number like $e^x$ so:
$$\alpha + \beta e^{x} + \gamma e^{2x} = 0$$
Suppose you have some solution for this with $\alpha$, $\beta$, $\gamma$ not all zero. Then, as you say
$$ \alpha + \beta + \gamma = 0\qquad \qquad (1)$$
Because this must be true at $x = 0$ but it must also be true at $x = \ln n$ which gives:
$$ \alpha + \beta n + \gamma n^2= 0\qquad \qquad (2)$$
for every $n > 1$.
It should be clear that this is unsolvable except when they are all zero. But to press the point I'll continue.
Substituting in $(1)$ gives $\alpha = -\beta - \gamma$, which we can plug into $(2)$ to get
$$ \beta (n-1) + \gamma (n^2 - 1)= 0$$
which must be true for all $n > 1$.
Now put, say, $n = 2$ and $n = 3$ to get the pair of equations:
$$ \beta + 3 \gamma = 0 \qquad 2\beta + 8\gamma = 0 $$
This solves for $\beta = \gamma = 0$.
So your functions are proved to be linearly independent.
