I'm stuck on this problem and would appreciate some help.
The set of all sequences on $\mathbb{R}$ constructs a vector space with the operations $(a_n)_{n\in \mathbb{N}} +(b_n)_{n\in \mathbb{N}}=(a_n+b_n)_{n\in \mathbb{N}}$ and $\lambda (a_n)_{n \in \mathbb{N}}= (\lambda a_n)_{n \in \mathbb{N}}$. Also real sequences that satisfy the equation $a_{n+2}=a_{n+1}+a_n$ construct a subspace $F$ of the aforementioned vector space of sequences.
a) Show that two sequences $(a_n)_{n \in \mathbb{N}}, (b_n)_{n\in \mathbb{N}}$ with $a_1=1,a_2=0$ and $b_1=0,b_2=1$ is a basis of $F$.
b) Show that two sequences $\left(\frac{1 +\sqrt5}{2}\right)^{n-1}$ and $\left(\frac{1 -\sqrt5}{2}\right)^{n-1}$ are also a basis of $F$.
c) Determine the Transition matrix between the bases in a) and the bases in b).
I think I have a acceptable proof for a) but I haven't been able to make any progress between b) and c). For part a) I start by noticing that subspace $F$ is just the set of all Fibonacci sequences and therefore $(a_n)_{n \in \mathbb{N}}, (b_n)_{n\in \mathbb{N}}$ is entirely dependent on its initial conditions in other words $(a_1,a_2),(b_1,b_2).$ As such there is an obvious correspondence between this and the standard basis of $\mathbb{R^2}$. Which means that
$x_{n+2}= \begin{pmatrix} x_{n+1} \\ x_n \end{pmatrix}$ then with $A= \begin{pmatrix} 1 &0 \\ 0&1 \end{pmatrix}$ it follows that $A \begin{pmatrix} x_{n+1} \\ x_n \end{pmatrix}=x_{n+2}$.
Best Answer
For b) you just have to verify that the two sequences are solutions of the problem and also linearly independent as you should have done in a). That will make you a basis if you already shown that the vector space is of dimension 2.
For c) you could explicit the general term of the sequences $ (a_n) $ and $(b_n)$ in the basis given in b) using the initial conditions.
Also be careful because $A\left(\begin{array}{c}x_{n+1} \\ x_{n}\end{array}\right)=x_{n+2}$ is not consistent, you have a vector on the left and a scalar on the right.