Proving that the set $u_1(x) = e^{a_1x}, \ldots, u_n(x) = e^{a_nx}$ is linearly independent.

Question

I'm following this proof from a book, but can't quite understand one of the steps. The proof in full is as such:

Suppose that, per the definition of linear independence that

$$\sum^n_{k = 1}c_ke^{a_kx} = 0$$

then, let $a_M$ be the biggest number in the set $\{a_1,\ldots,a_n\}$. Multiplying both sides of the above equality, we get

$$\sum^n_{k = 1}c_ke^{(a_k - a_M)x} = 0.$$

If $k \ne M$, then $a_k - a_m < 0$. By letting $ x\to+\infty$, we get $c_M = 0$. We can, then, eliminate the term $k = M$ from the given linear combination.

Finally, by successive applications of the same process, we get $c_k = 0$ for $k = 1, 2, \ldots, n$.

How can we know that $c_M = 0$ by letting $x$ tend to infinity? That's the only step I don't understand.

Answer 1

This is due to the fact that $\lim_{\rightarrow+\infty}e^{-x}=0$ by multiplyiing the equality by $e^{-a_M}x$ you obtain $c_M+\sum_{k\neq M}e^{(a_k-a_M)x}$, since $a_k-a_M<0$, $\lim_{x\rightarrow +\infty}e^{(a_k-a_M)x}=0$.

Answer 2

Here is another proof that uses the Vandermonde matrix and has the minor advantage of working with complex numbers as well:

Let $f(x) = \begin{bmatrix} e^{a_1x} & \cdots & e^{a_nx} \end{bmatrix}^T$, and note that $f^{(k)}(0) = \begin{bmatrix} a_1^k & \cdots & a_n^k \end{bmatrix}^T$. Then $V= \begin{bmatrix} f(0) & f^{(1)}(0) & \cdots & f^{(n-1)}(0) \end{bmatrix}$ is a Vandermonde matrix (https://en.wikipedia.org/wiki/Vandermonde_matrix) which is known to be invertible if the $a_k$ are distinct.

Hence if $c^T f(x) = 0$ for all $x$ we must have $c^TV = 0$ and hence $c=0$.