Fibonacci sequences as a vector space Can someone help me prove step by step that the the set V of all the Fibonacci sequences is a vector space?
I know I have to prove that such a set has the 10 properties relative to a vector space (closed under addition, scalar multiplication etc) but I am not familiar with Fibonacci sequences and I am struggling doing that.
 A: The set of Fibonacci type sequences  is a vector space.
Let $$ F_1 = a,b,a+b, a+2b, 2a+3b,......$$ and $$F_2 =  c,d,c+d, c+2d, 2c+3d,......$$ be two Fibonacci sequences, then we have   
$$ F_1 + F_2 = (a+c),(b+d), (a+b)+(c+d),( a+2b)+(c+2d),..... =$$
$$ (a+c),(b+d), (a+c)+(b+d),( a+c)+2(b+d),..... $$ is a Fibonacci sequence.
Similarly if $F$ is a Fibonacci sequence,  $\lambda F$ is also a Fibonacci sequence.
Other properties of a vector space are inherited from properties of sequences.
A: It is enough to prove that the null sequence is a Fibonacci sequence, that the sum of two Fibonacci sequences is a Fibonacci sequence and the product of a Fibonacci sequence by a scalar is a Fibonacci sequence. This proves the Fibonacci sequences form a subspace of the vector space of real sequences.
Btw what is a Fibonacci sequence. It is a sequence that follows the recursion, given $u_0$ and $u_1$:
$$u_{n+2}=u_{n+1}+u_{n}$$ 
Now if we take $u_0=u_1=0$ we get the null sequence. (Check it)
Now if $u$ and $v$ are Fibonacci one has
$$u_{n+2}=u_{n+1}+u_{n}\\u_{n+2}=u_{n+1}+u_{n}$$
Adding the two identities one gets
$$u_{n+2}+v_{n+2}=u_{n+1}+v_{n+1}+u_n+v_n$$
And this proves $u+v$ is Fibonacci. (Because $(u+v)_n=u_n+v_n$
Now multiplying one of the identities by $\lambda$, one gets
$$\lambda u_{n+2}=\lambda u_{n+1}+\lambda u_n$$
And this proves $\lambda u$ is Fibonacci. (Because $(\lambda u)_n=\lambda u_n$
A: Recall that Fibonacci sequence is definded by
$$F(n)=F(n-1)+F(n-2)$$
and it is determined by the initial conditions $F(0)=F_0$ and $F(1)=F_1$.
Then we need to check that


*

*zero sequence is a Fibonacci sequence and this is true by $F_0=F_1=0$

*$\forall a,b\in\mathbb{R}$ we have that $aF(n)+bG(n)$ is a Fibonacci sequence, that's true indeed


$$H(n)=F(n)+G(n)=F(n-1)+F(n-2)+G(n-1)+G(n-2)=\\=H(n-1)+H(n-2)$$
$$H(n)=aF(n)=a(F(n-1)+F(n-2))=aF(n-1)+aF(n-2)=\\=H(n-1)+H(n-2)$$
