Question

Prove that the set {e^ax ,e^bx is linearly independant if and only if a doesnt equal b

Answer 1

Two functions, considered as vectors, are linearly independent if there are no non-zero constants p and q such that
pe^ax + qe^bx = 0
where the 0 here is the zero function.

You need to show both ways. It's easier to show they are linearly dependent if and only if a is not equal to b. Try that.

Answer 2

kk Thanks James :D

Answer 3

i think there is a term called the wronskian which is just the determinant of the square matrix of the functions and their derivatives.

Answer 4

e^ax e^bx
a e^ax b e^bx

when b e^(a+b)x - a e^(a+b)x = 0 there are considered dependant; otherwise they are independant

Answer 5

so, if b e^(a+b)x = a e^(a+b)x ; and since e^() is never zero we can divide them off with no repurcussions

b = a results in dependance

Answer 6

yes that is what I was referring to :DDDD Thanks

Answer 7

You have to be careful with that result, as an everywhere zero Wronksian doesn't imply linear dependence.

You can show this directly without recourse to the Wronksian.

Answer 8

kk :DDD Thanks James

Answer 9

Might as well write this out. I'm going to show that the two functions are lin. dependent if and only if a=b. That then is equivalent to what you want to prove.

(=>): Suppose e^ax and e^bx as functions are linearly dependent. Then there exist constants p and q not equal to zero such that
p.e^ax + q.e^bx = 0, the zero function.
i.e.,
e^(a-b)x = q/p, a constant function

But the only way e^(a-b)x can be constant is if a-b = 0.

(<=): If a = b, then clearly
1.e^ax + (-1)e^bx = 0

qed

Answer 10

K thanks james I will write that down :D