How to verify the subspace of an infinite vector space is finite-dimensional? so on a homework for my linear algebra course, I am stuck with the following.
There is a subspace  $S=\{(x_1,x_2,...): x_n+x_{n+1}-x_{n+2}=0, \forall n \geq 1\} \subseteq \mathbb{F}^\infty$  My task is to answer if (a) S is finite dimensional, and then (b) what its dimension is.  My guess so far is to show that if $S$ is finite dimensional, then the list of vectors $(x_1,x_2,...)$ spans the set, which will generally contain vectors of the form $(v_1,v_2,...)$, for $\mathbb{F}^\infty$. Thus I somehow need to show that 
$\bigoplus_{i=1}^{\infty} c_ix_i=\sum_{j=1}^{\infty} v_j$ 
but this creates an infinite system of equations I don't know how to solve. Tips? Thank you
 A: Let $a$ and $b$ be any two numbers and consider the sequence $(x_n)_{n\in\Bbb N}$ such that:

*

*$x_1=a$;

*$x_2=b$;

*$(\forall n\in\Bbb N):x_{n+2}=x_n+x_{n+1}$.

Then $(x_n)_{n\in\Bbb N}\in S$ and every element of $S$ is of this type (with $a=x_1$ and $b=x_2$). So, the whole space $S$ depends upon the two parameters $a$ and $B$, which suggests that $\dim S=2$. Can you take it from here?
A: Notice that a sequence $\mathbf{x}:\mathbb{Z}_+\rightarrow\mathbb{R}$ is in $S$ satisfies
$$\begin{align}
\begin{pmatrix} x_{n+1} \\ x_n\end{pmatrix}&=
\begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}\begin{pmatrix} x_n\\ x_{n-1}\end{pmatrix}\tag{1}\label{one} \\
&=\begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}^{n-1}\begin{pmatrix} x_2\\ x_1\end{pmatrix}\tag{2}\label{two}\\
&=x_1\begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}^{n-1}\begin{pmatrix} 0 \\1\end{pmatrix} + x_2\begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix}^{n-1}\begin{pmatrix} 1 \\0\end{pmatrix}\tag{3}\label{three}
\end{align}
$$
for $n\geq 2$. Thus, by \eqref{one} each sequence $\mathbf{x}$ in $S$ is fully determined by the first two values $\mathbf{x}(1)=x_1$ and $\mathbf{x}(2)=x_2$.
By \eqref{two} and \eqref{three}, if $\mathbf{x}_1$ is the sequence with $\mathbf{x}_1(1)=1$ and $\mathbf{x}_1(2)=0$ and $\mathbf{x}_2$ is the sequence with $\mathbf{x}_2(1)=0$ and $\mathbf{x}_2(2)=1$, then the sequence $\mathbf{x}$ with $\mathbf{x}(1)=x_1$ and $\mathbf{x}(1)=x_2$ is given as
$$\mathbf{x}=x_1\mathbf{x}_1+ x_2\mathbf{x}_2$$
To complete you problem you may want to check that $\mathbf{x}_1$ and $\mathbf{x}_2$ are linearly independent.
