Linear independence Fibonacci vectors Let $F_n$ be the usual Fibonacci sequence i.e. with $F_0=0$, $F_1=1$ and $$F_n=F_{n-1}+F_{n-2}$$
I was wondering if the vectors $$\pmatrix{F_n \\ F_{n+1}} \text{ and } \pmatrix{F_{n+2} \\ F_{n+3}}$$
are linearly independent.
I took some random values and they were in fact linearly independent. So I strongly believe it is true. My first attempt to prove it was to show that if there exists $a,b \in \Bbb R$ such that $$a\pmatrix{F_n \\ F_{n+1}} +b \pmatrix{F_{n+2} \\ F_{n+3}}=0 \implies a=0,b=0$$
Rephrasing it gives that we want to show $$\pmatrix{a+b & b \\ b & a+2b} \pmatrix{F_{n} \\ F_{n+1}}=0 \implies a=0,b=0$$
for any $n\in \Bbb N$.
I really struggle to do it, so I thought of a recursion : if $\pmatrix{F_n \\ F_{n+1}} \text{ and } \pmatrix{F_{n+2} \\ F_{n+3}}$ (and all previous ones) are linearly independent then we would like to show that $$a\pmatrix{F_{n+1} \\ F_{n+2}} +b \pmatrix{F_{n+3} \\ F_{n+4}} $$
are linearly independent too but it does not go the good way when I try do it.
So is the result even true ? I am sure there is some shady property to show it very quickly.
 A: If $$v_n=\begin{pmatrix}F_n\\F_{n+1}\end{pmatrix}$$ it is easier to see that $v_n$ and $v_{n+1}$ are linearly independent. Then $v_{n+2}=v_n+v_{n+1}$ is also linearly independent to $v_n,$ due to the general rule:

If $u,v$ are independent, then $u$ and $u+v$ are independent.

So you only need to prove that $v_n,v_{n+1}$ are linearly independent.
But we can do that by induction on $n.$ Let $u=v_{n+1}, v=v_n,$ then, by the same result, $u=v_{n+1}$ and $v_{n+2}=u+v$ are independent.

Alternatively, show by induction:
$$\begin{pmatrix}0&1\\1&1\end{pmatrix}^n \begin{pmatrix}0&1\\1&2\end{pmatrix}= \begin{pmatrix}F_{n}&F_{n+2}\\F_{n+1}&F_{n+3}\end{pmatrix}$$ Show the left side is invertible, so the columns on the right side are independent.
A: Let $v_n = \begin{pmatrix} F_n \\ F_{n+1} \end{pmatrix}$. We have $v_{n+1} = Av_n$ where $A = \begin{pmatrix}0 & 1 \\ 1 & 1\end{pmatrix}$.
"$v_{n+2} = A^2 v_n$ and $v_n$ are linearly dependent" is equivalent to "$A^2 v_n = cv_n$ for some scalar $c$," which in turn is equivalent to "$v_n$ is an eigenvector of $A^2 = \begin{pmatrix}1 & 1 \\ 1 & 2\end{pmatrix}$." Since the eigenvectors of this matrix are not integer vectors, this is impossible.
