How to Prove the Linear Independence of Exponential Functions?

[phantomprime]

Lindemann–Weierstrass theorem??

Let X1, X2, X3,...Xn be an increasing sequence of real numbers. Prove that the n exponential functions e^x1t,e^x2t...e^xnt are linearly independent.

My question is how? I look at examples but how do I go about explaining this?

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Show that the set {1, sqrt(2), sqrt(3), sqrt(6)} is linearly independent over the rational numbers

 

[uman]

(sorry, ignore this post)

 

[maze]

I think if you just keep picking values of t, you will get a system of linear equations that is overdetermined (and unsolvable since the columns of rows of constants in the matrix are linearly independent).

 

[HallsofIvy]

Both of those look pretty straight forward from the definition of "independent".

Given a_1e^{x_1t}+ a_2e^{x_2t}+ \cdot\cdot\cdot+ a_ne^{x_nt}= 0
for all t, prove that a_1= a_2= \cdot\cdot\cdot= a_n= 0
I suspect it would be simplest to prove this by induction on n.

For the second one, you want to prove that if a+ b\sqrt{2}+ c\sqrt{3}+ d\sqrt{6}= 0, and a, b, c are rational numbers, then a= b= c= d= 0. You will, of course, use the fact that \sqrt{2} and \sqrt{3} are not rational.

 

[maze]

How would one go from the kth case to the k+1th?

 

[morphism]

If we multiply both sides of

a_1e^{x_1t}+ a_2e^{x_2t}+ \cdots + a_ne^{x_nt}= 0

by e^{-x_n t}, we get

a_1e^{(x_1 - x_n)t}+ a_2e^{(x_2 - x_n)t}+ \cdots + a_ne^{(x_{n-1} - x_n)t} + a_n = 0.

Now let t \to \infty.