Consider the vector space $C[0,1]$ and the following subset of $C[0,1]$: $$T = \{f_\lambda :[0,1]→\mathbb{R}\, | \, f_\lambda(x)=e^{\lambda x},λ∈\mathbb{R}\}$$
(that is, $T$ contains all the different exponential functions restricted to $[0, 1]$).
Note that, for every $λ_1,λ_2 \in \mathbb{R}$, if $λ_1\neq λ_2$, then $e^{λ_1} \neq e^{λ_2}$ ⇒ $f_{λ_1}(1) \neq f_{λ_2}(1) ⇒ f_{λ_1} \neq f_{λ_2}$. In other words, the cardinality of the set $T$ is the same as the cardinality of $\mathbb{R}$.
Prove that T is a linearly independent subset of $C[0,1]$.
Proof attempt:
Since the set T is clearly infinite, thus to show linear independence we need to prove every finite subset of T is linearly independent.
Therefore we take n elements {${e^{λ_1x_1},…,e^{λ_nx_n}}$} where $x \in [0,1]$
then we take scalars $a_1….a_n$ and set ${a_1e^{λ_1x_1}+…+a_ne^{λ_nx_n}}=0$
So here is where I get stuck/fuzzy, I believe this should be the end of the proof since there is no way the exponential function adds up to equal $0$ unless the scalars are equal to $0$ but I'm unsure.
corrections and critiques would be appreciated
Best Answer
Do you mean exponential functions rather than exponential values? If functions, I think if you consider $\{e^{λ_1x},...,e^{λ_nx}\}$ with all $\lambda_i$ distinct you can prove that this is linearly independent by induction.
It is true for $n=1$, clearly.
If $n=2$, consider $f(x) =a_1e^{b_1x}+a_2e^{b_2x} $. If $f(x) = 0$ then, dividing by $e^{b_1x}$, $0 =a_1+a_2e^{(b_2-b_1)x} $, and this can be true only if $b_1 = b_2$.
The induction step for $f(x) =\sum_{k=1}^n a_ke^{b_kx} $ can be done by noting that if $f(x) = 0$ then $0 =g(x) =f(x)e^{-b_1x} =\sum_{k=1}^n a_ke^{(b_k-b_1)x} =a_1+\sum_{k=2}^n a_ke^{(b_k-b_1)x} $ so that $0 =g'(x) =\sum_{k=2}^n a_k(b_k-b_1)e^{(b_k-b_1)x} $ but this can not be zero by the induction hypothesis.