How do I show that the set $\{ e^x , ... ,e^{nx} \}$ is linearly independent?
I tried using induction as the base case of $\{ e^x \}$ and even $\{ e^x, e^{2x} \}$ is easy, but I can't use the I.H. to go further. What I try to do with the induction step is plug in values of x and try to force the last coefficient to be 0, to no avail.
To show that the set is LI, you need to show that $$\sum_{k=1}^n a_ke^{kx}=0\Rightarrow a_k=0,k=1,2,\cdots,n\quad\forall x$$ Then take $n$ values of $x=x_1,\cdots ,x_n$, form a set of $n$ equations and then the requirement needs to show that the equation $$Xa=0$$ has only the trivial solution $a=0$ where $$X=\begin{pmatrix} 1 & e^{x_1} & \cdots e^{nx_1}\\ \vdots & \vdots & \vdots \\ 1 & e^{x_n} & \cdots e^{nx_n} \end{pmatrix}$$ and $a=(a_1\cdots a_n)^T$ Since $X$ is a Vandermonde matrix, this is evident. Hence, the set is LI.
