Linear Independence 2De260

1. **Problem Statement:** Prove that the set $\{e^{ax}, e^{bx}\}$ is linearly independent on $\mathbb{R}$. 2. **Definition:** A set of functions $\{f_1, f_2\}$ is linearly independent if the only solution to the equation $c_1 f_1(x) + c_2 f_2(x) = 0$ for all $x$ in $\mathbb{R}$ is $c_1 = 0$ and $c_2 = 0$. 3. **Set up the equation:** Suppose there exist constants $c_1, c_2$ such that $$c_1 e^{ax} + c_2 e^{bx} = 0 \quad \text{for all } x \in \mathbb{R}.$$ 4. **Goal:** Show that $c_1 = 0$ and $c_2 = 0$ are the only solutions. 5. **Case 1: $a = b$** If $a = b$, then the set is $\{e^{ax}, e^{ax}\}$ which is not linearly independent because the two functions are the same. 6. **Case 2: $a \neq b$** Rewrite the equation as $$c_1 e^{ax} = -c_2 e^{bx}.$$ Divide both sides by $e^{bx}$ (which is never zero): $$c_1 e^{ax} \div e^{bx} = -c_2.$$ Using properties of exponents: $$c_1 e^{(a-b)x} = -c_2.$$ 7. **Analyze the equation:** The left side $c_1 e^{(a-b)x}$ is a function of $x$, while the right side $-c_2$ is a constant. For this equality to hold for all $x$, the function on the left must be constant. 8. **Since $a \neq b$, $e^{(a-b)x}$ is not constant.** Therefore, the only way for $c_1 e^{(a-b)x} = -c_2$ to hold for all $x$ is if $c_1 = 0$. 9. **Substitute $c_1 = 0$ back into the original equation:** $$0 + c_2 e^{bx} = 0 \quad \Rightarrow \quad c_2 e^{bx} = 0.$$ Since $e^{bx} \neq 0$ for all $x$, it follows that $c_2 = 0$. 10. **Conclusion:** The only solution is $c_1 = 0$ and $c_2 = 0$, so the set $\{e^{ax}, e^{bx}\}$ is linearly independent on $\mathbb{R}$ when $a \neq b$.