1

Prove that $\{\mathbb e^{r_1x}, \mathbb e^{r_2x}..., \mathbb e^{r_nx}\}$ are linearly independent where $r_1,r_2,...r_n$ are not equal.

now I know that this question has been asked before but it's not the same this doesn't end up to be a Vandermonde matrix.
so is there a workaround this as well.

hamid kamali
  • 3,201
Moutasem
  • 21
  • 4
    Possibly duplicate of http://math.stackexchange.com/questions/23139/proof-of-linear-independence-of-eat or http://math.stackexchange.com/questions/633724/prove-ex-e2x-enx-is-linear-independent-on-the-vector-space-of-ma. – Santiago Mar 31 '16 at 10:50
  • 1
    Why do you think this does not lead us a Vandermonde matrix? – crbah Mar 31 '16 at 10:59
  • There are four answers in the other question. Only one of them uses Vandermonde matrix. – user1551 Mar 31 '16 at 11:27

2 Answers2

3

By re-indexing, we can assume $r_{1} > r_{2} > ... > r_{n}$. Suppose there are $c_{i}$ not all equal zero such that $\sum_{i=1}^{n}c_{i}e^{r_{i}x} =0$. Let $k$ be the smallest index such that $c_{k} \neq 0$. Then $\sum_{i=1}^{n}c_{i}e^{r_{i}x} =0 \implies \sum_{i=k}^{n}c_{i}e^{r_{i}x} =0$.

Then on multiplying both sides of this shorter sum by $e^{-r_{k}x}$, we get $\sum_{i=k}^{n}c_{i}e^{(r_{i} - r_{k})x} = 0$. Since the LHS is a continuous function, we take the limit as $x \to \infty$ and we get $c_{k} = 0$, which is a contradiction.

Daniel Akech Thiong
  • 4,061
3

Say $c_1.e^{r_1 x} + c_2.e^{r_2 x} + ... + c_n.e^{r_n x} = 0$

Differentiate w.r.t. x:

$r_1.c_1.e^{r_1 x} + r_2.c_2.e^{r_2 x} + ... + r_n.c_n.e^{r_n x} = 0$

n-th order derivative:

$r_1^n.c_1.e^{r_1 x} + r_2^n.c_2.e^{r_2 x} + ... + r_n^n.c_n.e^{r_n x} = 0$

Now we have

$\begin{bmatrix} 1 & 1 &... & 1\\ r_1 & r_2 &... & r_n\\ . & . &... & . \\ . & . &... & . \\ r_1^n & r_2^n &... & r_n^n\end{bmatrix}$ . $\begin{bmatrix} c_1.e^{r_1 x} \\ . \\ . \\ c_n.e^{r_n x} \end{bmatrix}$ = $0$.

The matrix is Vander Monde matrix, so it is invertible, thus $\begin{bmatrix} c_1.e^{r_1 x} \\ . \\ . \\ c_n.e^{r_n x} \end{bmatrix} = 0$.

That is each $c_k.e^{r_k x} = 0$, which means $c_k = 0$.

crbah
  • 1,162