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$(\mathbb{R}^{\mathbb{R}},+, \cdot)$ is an $\mathbb{R}$ Vector space. For all $n \in \mathbb{N}^*$, prove that the set $\left(e^{ kx }\right)_{k \in [|1,n|]}$ is linearly independent.

I have already proved it by induction, but I'm trying to prove it now using isomorphisms, and here's how I've gone through it so far. Let $n \in \mathbb{N}^*$, and let us prove that the set $\left(e^{ kx }\right)_{k \in [|1,n|]}$ is linearly independent. Consider the following morphism $$ \begin{align} f:\:(\mathbb{K}_{n}[X],+,\cdot)& \longrightarrow (f(\mathbb{K}_{n}[X]),+,\cdot) \\ &P \mapsto \tilde{P} \circ(e^{ x }) \end{align} $$ Since the set $(X^{k})_{k \in [|1,n|]}$ is already linearly independent in $\mathbb{K}_{n}[X]$, all I gotta do now is prove that $f$ is injective to deduct that $\left(e^{ kx }\right)_{k \in [|1,n|]}$ is linearly independent in $\mathbb{R}^{\mathbb{R}}$. Now here is where im stuck, as I find it quite difficult to prove that $f$ is injective. I wanna see if it is possible, in some way, to prove that $f$ is injective.

J. W. Tanner
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Sewshley
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    Years ago, I wrote this answer to treat this kind of problems. Hopefully it is usefult for you – Giuseppe Negro Mar 28 '24 at 16:23
  • Showing injectivity of $f$ could look like taking an arbitrary $P=\sum_{k=1}^n a_kX^k$ and showing that if $f(P)=0$, then each $a_k=0$. But this is just showing that $\sum_{k=1}^n a_ke^{kx}\equiv 0$ implies that each $a_k=0$, which is the original problem again. I don't see any way that you can prove that $f$ is injective which can't essentially work for the original problem. – user469053 Mar 28 '24 at 16:31
  • There could be alternatives (for example, showing that the dimension of the range is $n$ and using the rank nullity theorem), but I suspect that when you try to identify the range of $f$, you end up with the original problem again. – user469053 Mar 28 '24 at 16:32
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    I do think there is an alternative proof strategy here: Rather than using the fact that $(X^k)$ is already known to be linearly independent and using $f$, one could attempt to prove that $(e^{kx})$ is linearly independent using the same method of proof that was used for $(X^k)$. – user469053 Mar 28 '24 at 16:33
  • @user469053 That is exactly what I stumble upon. I guess proving the injectivity of $f$ is equivalent to solving the problem itself. – Sewshley Mar 28 '24 at 16:43
  • @GiuseppeNegro I like your idea. As a matter of fact, it makes it very easy to prove that a set of functions is linearly indepdant (given the functions) that way. – Sewshley Mar 28 '24 at 16:52

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