# Proof, linear independence

• Jun 4th 2011, 10:32 AM
I-Think
Proof, linear independence
Let $\displaystyle f,g,\in{F(R,R)}$ be the functions defined by $\displaystyle f(t)=e^{rt}$and $\displaystyle g(t)=e^{st}$, where $\displaystyle r\neq{s}$. Prove that $\displaystyle f$and $\displaystyle g$are linearly independent in $\displaystyle F(R,R)$.

Assume $\displaystyle ae^{rt}+ be^{st}=0$
$\displaystyle ae^{rt} = -be^{st}$
$\displaystyle ln(a)+rt = ln(-b) +st$
$\displaystyle ln(\frac{-a}{b})=t(s-r)$

For our assumption to be true, the values of $\displaystyle a$ and $\displaystyle b$ are dependent on the value of $\displaystyle t$, therefore there is no universal value of $\displaystyle a$ or $\displaystyle b$, except $\displaystyle 0$, for which $\displaystyle af+bg=0$
Thus, these functions are linearly independent

This proof is correct?
• Jun 4th 2011, 10:49 AM
girdav
You don't know if $\displaystyle a$ or $\displaystyle b$ are positive. But you can take $\displaystyle t=0$ in the equation, go back to the equation and take the derivative on both side, then take $\displaystyle t=0$.
• Jun 4th 2011, 10:59 AM
I-Think
Actually, we do know that one of the two constants is negative and the other is positive. $\displaystyle t\in{R}$ and $\displaystyle e^x$ is never negative, so for the equation to be true, one of the constants must be negative and the other positive.
• Jun 4th 2011, 11:22 AM
girdav
Yes, it would be perfect if you add "We assume $\displaystyle a>0$, if it's not the case we switch $\displaystyle r$ and $\displaystyle s$.
• Jun 4th 2011, 12:51 PM
Deveno
$\displaystyle ae^{rt}+be^{st}$ is the 0 function, not just the number 0. this means this equation holds for any and all values of t.

in particular, it holds for t = 0, whence b = -a.

thus we have $\displaystyle a(e^{rt} - e^{st}) = 0$ for all t. if a is not 0, then

$\displaystyle e^{rt} = e^{st}$ for all real t, taking logs of both sides gives r = s, a contradiction, so a = 0 and b = -a = 0.

EDIT: why do i do it this way? because in the original poster's proof, he takes the log of a number which he is trying to prove is 0.

now, by the previous two comments, you can actually assume that a > 0, and b < 0, and then the original poster's argument works,

but it is not easy to see that ln(a/-b) = t(s - r) immediately implies a contradiction. i feel it is more straight-forward to use the fact that R

is a field (and thus has no zero divisors).