Linear Algebra: Linear Combinations

lockedup

1. The problem statement, all variables and given/known data
Let V = {f: [tex]\mathbb {R}\rightarrow\mathbb {R}[/tex]} be the vector space of functions. Are f₁ = e^x, f₂ = e^-x (both [tex]\in[/tex] V) linearly independent?


2. Relevant equations
0 = ae^x + be^-x Does a = b = 0?


3. The attempt at a solution
My first try, I put a = e^-x and b = -e^x. He handed it back and told me to try again. I think the problem was that my a and b were not constants. But how to prove that there are no constants that will make the equation 0? I wrote some stuff down about the fact that, if a=0, then b = 0 (and the converse). Is that sufficient or am I way off?

nicksauce

One way to do it would be a proof by contradiction.

Suppose there are constants, not both 0, a and b, such that aexp(x) + bexp(-x) = 0 for all x. Then aexp(x) = -bexp(-x), so -a/b * exp(2x) = 1 for all x. I won't complete it for you, but look at x=0, and see what restriction it places on a/b. Then look at a different point and you will reach a contradiction. (Also note, I've implicitly assumed b is nonzero, so you should handle that case as well)

lanedance

how about

a.e^x+b.e^-x = 0
multiply by e^-x
a.e^2x+b = 0

this is not true in general for all x in the reals unless a=b=0

Mark44

Yes, a = b = 0 is one solution, and is always a solution regardless of whether these functions are linearly dependent or linearly independent. The real question is whether this solution, the trivial solution, is the only solution. If so, the functions are linearly independent. If not, they are linearly dependent.