# linear algebra : show that $~f~$ is a linear map.

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 $$\operatorname{Im}(f) \supseteq \mathbb{R} \times \mathbb{R}\;?$$

• 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. Jul 20, 2019 at 16:15
• I've edited your question to include what I believe you're trying to ask. Jul 20, 2019 at 22:49
• 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 Jul 20, 2019 at 22:55
• 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 $(U_n)_{n\in\Bbb N}\in F$ .
– Surb
Jul 20, 2019 at 22:55
• By the way, the "vector space of Fibonacci sequences" is very cool :) never thought about such spaces.
– Surb
Jul 20, 2019 at 22:57