
#1
Mar309, 12:46 AM

P: 70

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 f_{1} = e^{x}, f_{2} = 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? 



#2
Mar309, 01:04 AM

Sci Advisor
HW Helper
P: 1,276

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) 



#3
Mar309, 01:07 AM

HW Helper
P: 3,309

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 



#4
Mar309, 09:55 AM

Mentor
P: 21,067

Linear Algebra: Linear Combinations 


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