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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

1 Answers1

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Since the set \begin{equation}( (X^k)_{k \in [|1,n|]} )\end{equation} is already linearly independent in \begin{equation}( \mathbb{K}_n[X] )\end{equation}, to show that \begin{equation}( f )\end{equation} is injective, we need to prove that the kernel of \begin{equation}( f )\end{equation} is trivial, i.e., the only polynomial that maps to the zero function under \begin{equation}( f )\end{equation} is the zero polynomial.

Suppose \begin{equation}( f(P) = 0 )\end{equation} for some non-zero polynomial \begin{equation}( P \in \mathbb{K}_n[X] )\end{equation}. This means that \begin{equation}( P(e^x) = 0 )\end{equation} for all \begin{equation}( x \in \mathbb{R} )\end{equation}. However, the exponential function \begin{equation}( e^x )\end{equation} is never zero for any real number ( x ), and a non-zero polynomial can have at most a finite number of roots, which is a contradiction. Hence, the only polynomial that can map to the zero function is the zero polynomial itself, and therefore \begin{equation}( f )\end{equation} must be injective.

It follows that the set \begin{equation}( \left(e^{kx}\right)_{k \in [|1,n|]} )\end{equation} is linearly independent in \begin{equation}( \mathbb{R}^{\mathbb{R}} )\end{equation} because the image of a linearly independent set under an injective linear transformation is also linearly independent. I apologize for any confusion. Let's clarify the logic behind the proof.

The intent was to leverage the properties of the exponential function and polynomials to establish the injectivity of the mapping \begin{equation}( f )\end{equation}, which is defined as:

\begin{equation}[ f(P) = P(e^x) ]\end{equation}

where \begin{equation}( P )\end{equation} is a polynomial in \begin{equation}( \mathbb{K}_n[X] )\end{equation}.

Premise 1: The exponential function \begin{equation}( e^x )\end{equation} is never zero for any \begin{equation}( x \in \mathbb{R} )\end{equation}.

Premise 2: A non-zero polynomial \begin{equation}( P \in \mathbb{K}_n[X] )\end{equation} can have at most a finite number of roots.

Claim to Prove (Injectivity of \begin{equation}( f ))\end{equation}: If \begin{equation}( f(P) = P(e^x) = 0 )\end{equation} for all \begin{equation}( x \in \mathbb{R} )\end{equation}, then \begin{equation}( P )\end{equation} must be the zero polynomial.

Here's the breakdown of the argument:

  1. If \begin{equation}( f(P) = P(e^x) )\end{equation} is the zero function, it means that \begin{equation}( P(e^x) = 0 )\end{equation} for all \begin{equation}( x \in \mathbb{R} )\end{equation}.

  2. The exponential function \begin{equation}( e^x )\end{equation} maps every real number to a positive real number and is surjective onto the positive reals. Therefore, if \begin{equation}( P(e^x) = 0 )\end{equation} for all \begin{equation}( x \in \mathbb{R} )\end{equation}, then \begin{equation}( P(y) = 0 )\end{equation} for all positive \begin{equation}( y \in \mathbb{R} )\end{equation}.

  3. Since \begin{equation}( e^x )\end{equation} covers all positive real numbers, \begin{equation}( P )\end{equation} must have infinitely many roots if \begin{equation}( P )\end{equation} is not the zero polynomial, contradicting Premise 2. The only polynomial that has an infinite number of roots is the zero polynomial.

Thus, for \begin{equation}( P(e^x) )\end{equation} to be identically zero, \begin{equation}( P )\end{equation} itself must be the zero polynomial. This means that the only element in the kernel of \begin{equation}( f )\end{equation} is the zero polynomial, and thus \begin{equation}( f )\end{equation} is injective.

The commenter might have misunderstood the application of Premise 2 or the scope of the exponential function's surjectivity onto the positive reals. To be clear, the argument does not claim that \begin{equation}( e^x = 0 )\end{equation} has any solutions; rather, it uses the fact that \begin{equation}( e^x )\end{equation} takes on every positive value to argue that a polynomial that maps every output of \begin{equation}( e^x )\end{equation} to zero must itself be the zero polynomial.

Saradamani
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  • Perfect! I've completely forgotten about using the Kernel. Thank you so much!! – Sewshley Mar 28 '24 at 16:54
  • @Sewshley This answer is not correct. – user469053 Mar 28 '24 at 17:18
  • @user469053 The proof looks fine to me. What's incorrect about it? – Sewshley Mar 28 '24 at 17:25
  • The answer uses the following syllogism: Premise $1.$ $e^x=0$ has no solutions. Premise $2$. Any non-zero polynomial has only finitely many roots. Conclusion$(*)$: $P(e^x)\equiv 0$ is a contradiction.

    These two premises do not lead to Conclusion$(*)$. I'm not even sure why the answerer thinks these two premises lead to Conclusion$(*)$.

     – user469053 Mar 28 '24 at 23:58