What is the basis for $F$ in linear algebra?

[Dethrone]

Let $F$ be the set of infinite sequences $(a_1,a_2,a_3...)$, where $a_i \in \Bbb{R}$ that satisfy
$a_{i+3}=a_i+a_{i+1}+a_{i+2}$
This describes a finite-dimensional vector space. Determine a basis for $F$.

 

[Nono713]

Spoiler

It is easy to see that any sequence $(a_i)$ is uniquely determined by its first three elements, so this vector space has dimension 3. A suitable basis is the set of the three sequences $x, y, z$ defined by:
$$x = (1, 0, 0, 1, 1, 2, 4, \cdots)$$
$$y = (0, 1, 0, 1, 2, 3, 6, \cdots)$$
$$z = (0, 0, 1, 1, 2, 4, 7, \cdots)$$
i.e. $x$, $y$ and $z$ are the sequences defined by the first three elements $(1, 0, 0)$, $(0, 1, 0)$ and $(0, 0, 1)$ respectively. It's easy to see that the basis $\{ x, y, z \}$ is linearly independent, for suppose there exists $\lambda_1, \lambda_2, \lambda_3 \in \mathbb{R}$ such that:
$$\lambda_1 x + \lambda_2 y + \lambda_3 z = 0$$
Where $0$ is of course the zero sequence. But that would imply:
$$\lambda_1 (1, 0, 0, \cdots) + \lambda_2 (0, 1, 0, \cdots) + \lambda_3 (0, 0, 1, \cdots) = (0, 0, 0, \cdots)$$
That is:
$$(\lambda_1, 0, 0, \cdots) + (0, \lambda_2, 0, \cdots) + (0, 0, \lambda_3, \cdots) = (0, 0, 0, \cdots)$$
In other words, $\lambda_1 = \lambda_2 = \lambda_3 = 0$ and so this set is linearly independent. Finally, it's easy to see that this set spans the entire vector space, since every sequence $(a_i)$ with first three elements $a_1, a_2, a_3$ can be written as:
$$a_1 x + a_2 y + a_3 z$$
Which is in the vector space and has its first three elements equal to $a_1, a_2, a_3$ and so must be equal to $(a_i)$. Therefore $\{ x, y, z \}$ is a basis of this vector space.

 

[I like Serena]

Spoiler

I hope you'll forgive me that I'm using isomorphisms of vector spaces. (Blush)

Let $f$ be the function $F \to \mathbb R^3$ given by $(a_1,a_2,a_3, ...) \mapsto (a_1,a_2,a_3)$.
Since all elements following $a_3$ are uniquely determined by $a_1,a_2,a_3$, $f$ is a bijection.
Moreover, since $F$ is a vector space, it follows that for all $x,y \in F, \lambda \in \mathbb R$ we have: $f(x+y)=f(x)+f(y)$ and $f(\lambda x) = \lambda f(x)$.
Thus $f$ is an isomorphism of vector spaces.

Since {(1,0,0), (0,1,0), (0,0,1)} is a basis for $\mathbb R^3$, it follows that {(1,0,0,...), (0,1,0,...), (0,0,1,...)} is a basis for $F$. (Nerd)