Show linear independence Is the Set $$S=\{e^{2x},e^{3x}\}$$ linearly independent?? And answer says Linearly independent over any interval $(a,b)$,only when $0$ doesnot belong to $(a,b)$
How do I proceed??
Thanks for the help!!
 A: I assume that you're working on the vector space of continuous real maps over the interval $(a,b)$.
So let's consider the two functions $f(x)=e^{2x}$ and $g(x)=e^{3x}$ and suppose that
$$
\alpha f+\beta g = 0.
$$
This means that, for every $x\in(a,b)$, we have
$$
\alpha f(x)+\beta g(x)=0
$$
and, in particular, for $x=c$ and $x=d$, where we assume $a<c<d<b$ (which is possible whenever $a<b$ (which may also be infinity); then
$$
\begin{cases}
\alpha e^{2c}+\beta e^{3c}=0\\
\alpha e^{2d}+\beta e^{3d}=0
\end{cases}
$$
We can divide the first equation by $e^{2c}\ne0$ and the second by $e^{2d}$ getting
$$
\begin{cases}
\alpha+\beta e^{c}=0\\
\alpha+\beta e^{d}=0
\end{cases}
$$
Subtract the first from the second to get
$$
\beta(e^d-e^c)=0
$$
Since $c<d$, we have $e^d\ne e^c$, so we conclude $\beta=0$ and, substituting in the first equation, also $\alpha=0$.
So the two functions are linearly independent no matter whether $0\in(a,b)$ or not.
A: Suppose $\;r,s\in\Bbb R\;$ are such that
$$re^{2x}+se^{3x}=0\;\;,\;\;\forall\,x\in (a,b)\implies e^{2x}(r+se^x)=0$$
since $\;e^t\neq0\;\;\forall\,t\in\Bbb R\;$ , we get that
$$r+se^x=0\iff e^x=-\frac rs\;\;\forall\;x\in (a,b)\in\Bbb R$$
and since the exponential is not a constant function on any non-trivial interval $\;(a,b)\;$ we deduce that it must $\;r=s=0\;$
A: Consider $$c_1e^{2x}+c_2e^{3x}=0$$. Then Differentiating once you will get $$2c_1e^{2x}+3c_2e^{3x}=0$$. Again differentiate it to get $$4c_1e^{2x}+9c_2e^{3x}=0$$. Plug in a value $x_0$ such that $x_0$ belongs to $(a,b)$. Then the equation reduces to 
$$2c_1e^{2x_0}+3c_2e^{3x_0}=0$$ and $$4c_1e^{2x_0}+9c_2e^{3x_0}=0$$. This is a linear system of equations with determinant = $6e^{5x_0}$ which is never zero. Hence $c_1=c_2=0$.
A: Hint: Consider the Wronskian. 
For more details see http://en.wikipedia.org/wiki/Wronskian
