We say that a sequence $(U_n)_{n \in \Bbb N} \subset \mathbb{R}$ is Fibonacci if it satisfies $\ U_{n+2} = U_{n+1} + U_n, \ \forall n \in \mathbb{N}$. Let $F$ be the set of all Fibonacci sequences.

We have the function $f: F \to \mathbb{R} \times \mathbb{R}$ that exists: $f(U_n) = (U_0,U_1)$

We must demonstrate that $f$ is a linear isomorphism between $F$ and $\Bbb R \times \Bbb R$.

It's it easy to show that $f$ is a linear map.
In order to prove that it is an isomorphism, we must demonstrate also that $f$ is one-to-one and and onto. For the injectivity of $f$, we have $f$ is a linear map so all we have to show is that $\operatorname{ker}(f) = \{0 \}$.
Let $U_n \in \operatorname{ker}(f)$: then $$ f(U_n) =(0,0) = (U_0,U_1) $$ since if $n=2:\ U_2 = 0 + 0$ and by double recurrence: $U_{n+1} = U_n = 0$ - so it's easy to show that $U_{n+2} =U_{n+1}= 0$ (the hypothesis of the exercise). Now, since $U_n = 0$ we have $\operatorname{ker}(f)= \{0\}$ then $f$ is one to one.
To prove the surjectivity of $f$ we must show that $\operatorname{Im}(f) = \mathbb{R} \times \mathbb{R}$: now the inclusion $\operatorname{Im}(f) \subseteq \mathbb{R} \times \mathbb{R}$ is trivially true but the second is not. I spend so much time without getting nothing: in sum, how can I prove that $$ \operatorname{Im}(f) \supseteq \mathbb{R} \times \mathbb{R}\;? $$

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    $\begingroup$ The only definition of $f$ that you give is $f(U_n) = (U_0,U_1)$. This is certainly not a linear map unless $U_0=U_1=0$, so I think you must have made a mistake. $\endgroup$
    – saulspatz
    Jul 20, 2019 at 16:15
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    $\begingroup$ I've edited your question to include what I believe you're trying to ask. $\endgroup$ Jul 20, 2019 at 22:49
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    $\begingroup$ To show that $f$ is surjective, it suffices to state the following: for any "initial conditions" $(x_1,x_2) \in \Bbb R \times \Bbb R$, there exists a Fibonacci sequence $(U_n)$ that satisfies $U_0 = x_1$ and $U_1 = x_2$. I would say that this holds trivial by the nature of the recurrence, so I'm not sure what it is we should say $\endgroup$ Jul 20, 2019 at 22:55
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    $\begingroup$ For surjectivity: take any $(x,y)\in \Bbb R \times \Bbb R$ and let $(U_n)_{n\in\Bbb N}\subset \Bbb R$ be defined as $U_0=x,U_1=y$ and $U_{n+2}=U_{n+1}+U_n$, then $f((U_n)_{n\in\Bbb N})=(x,y)$ and $(U_n)_{n\in\Bbb N}\in F$ . $\endgroup$
    – Surb
    Jul 20, 2019 at 22:55
  • $\begingroup$ By the way, the "vector space of Fibonacci sequences" is very cool :) never thought about such spaces. $\endgroup$
    – Surb
    Jul 20, 2019 at 22:57


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