Show That $e^x$ and $e^{-x}$ are linearly independent

Question

To test for linear independence, let $$c_1e^x+c_2e^{-x}=0$$ Where $c_1$ and $c_2$ must equal $0$ for the two functions to be demonstrated to be linearly independent.

So for $x=0$: $$c_1=-c_2$$And, as $x$ approaches $-\infty$: $$c_1(0)+c_2(\infty)= 0$$ Then $c_2$ must equal 0, and:$$c_2 = 0 = c_1$$

Is this sufficient to demonstrate linear independence?

Answer 1

Let $x$ be another value. For example, let $x=1$ and you will have $$c_1e+c_2e^{-1}=0$$

with that you can solve for $c_1$ and $c_2$ uniquely and they are zero or show the determinant of the corresponding matrix is non-zero.

$$\begin{bmatrix}e & e^{-1} \\ 1 & 1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2\end{bmatrix}= \begin{bmatrix} 0 \\ 0\end{bmatrix}$$

Answer 2

For $x=Ln(2)$, $2c_1+{1\over 2}c_2=0$, with $c_1+c_2=0$ you deduce the result.

Answer 3

Wronskian of $e^x$ and $e^{-x}$:

$$\begin{vmatrix} e^x&e^{-x}\\ e^x&-e^{-x}\\\end{vmatrix}=-2\neq 0$$so.....?

Answer 4

If $c_1e^x + c_2e^{-x} = 0,$ then $c_1e^{2x} + c_2 =0.$ Assuming $c_1\ne 0,$ we then get $e^{2x}=-c_2/c_1,$ which is a contradiction. Thus $c_1=0,$ and therefore $c_2=0.$

Answer 5

Your work doesn’t make sense because $c_2 \infty$ is meaningless. Instead note that,

$$c_1 e^x+c_2 e^{-x}=0$$

Can be rewritten as,

$$c_1+c_2 e^{-2x}=0$$

Simply by dividing both sides by $e^x$. Since the equality holds for all $x$, it must hold as $x \to \infty$.

This gives,

$$c_1=0$$

A similar trick but with division by $e^{-x}$ and letting $x \to -\infty$, gives $c_2=0$.