Proof, linear independence

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Jun 4th 2011, 10:32 AM
I-Think

Proof, linear independence

Let $f,g,\in{F(R,R)}$ be the functions defined by $f(t)=e^{rt}$ and $g(t)=e^{st}$ , where $r\neq{s}$ . Prove that $f$ and $g$ are linearly independent in $F(R,R)$ .

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

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

This proof is correct?
Jun 4th 2011, 10:49 AM
girdav

You don't know if $a$ or $b$ are positive. But you can take $t=0$ in the equation, go back to the equation and take the derivative on both side, then take $t=0$ .
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. $t\in{R}$ and $e^x$ 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 $a>0$ , if it's not the case we switch $r$ and $s$ .
Jun 4th 2011, 12:51 PM
Deveno

$ae^{rt}+be^{st}$ 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 $a(e^{rt} - e^{st}) = 0$ for all t. if a is not 0, then

$e^{rt} = e^{st}$ 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).