Let $B:=A^2-A-id$, where $A$ is the left shift operator. Find $dim(\ker(B))$, basis for $\ker(B)$ and if $B$ is surjective. Let $V$ be the infinite dimensional $\mathbb{R}$-vector space of sequence of real numbers $(a_0, a_2, ...)$. Let $A\in \operatorname{End}(V)$ (endomorphism) be the left shift operator, so that $(a_1, a_2, ...) \to (a_2, a_3, ...) $ and define:
\begin{equation}
B:=A^2-A-\text{id} \in \operatorname{End}(V)
\end{equation}
Show that B is surjective. Determine also the dimension of $\ker(B)$ and find a Basis of $\ker(B)$.

I am confused about the concepts of surjectivity (particularly, how to prove it), how to express kernels formally, and how to build basis. My approach to the given problem was the following:
a) As $$B(a_1, a_2, ...) = (a_3, a_4, ...) - (a_2, a_3, ...) - (a_1, a_2, ...) = (a_3-a_2-a_1, a_4-a_3-a_2, ...)$$
I think it is surjective, as all elements from $\operatorname{img}(B)$ are the result of $B(a_1, a_2, ...)$, but how should I express or prove this in a more formal way? (please, correct me if I am wrong)
b) First, I want to express $B(a_1, a_2, ...)$ as follows (am I right to do it this way?):
$$Ba = (a_{k+2}-a_{k+1}-a_{k}|k\in \mathbb{N}\text{ and }a_k\in\mathbb{R})$$.
I suppose that I can express the kernel as follows:
$$\ker(B)=(x_k|x_{k+2}=x_{k+1}+x_{k})$$
As each $x_k$ is determined by another two elements, then follows $\dim(\ker(B))=2$. Am I right? I also think there should be a better way to express this statement (or to calculate the dimension in general).
c) How should I calculate the Basis of $\ker(B)$?
Thank you for your help.
 A: Let me adress surjectivity first. You start with a sequence $b=(b_1, b_2, \dots)$ and you want to find some $a=(a_1, a_2, \dots)$ such that $B(a)=b$. Now we can express this in terms of equations for every entry. Namely, for the first entry we get
$$ b_1 =a_3 - a_2 - a_1.$$
We can for example pick $a_1=0=a_2$ and $a_3=b_1$. For the second entry we get
$$ b_2 = a_4 - a_3 - a_2.  $$
With our previous choice we get
$$ b_2 = a_4 - b_1 $$
and hence, we have to choose $a_4= b_1+b_2$. Now we could go on. We just recursively define
$$ a_{n+2} = b_n + a_{n+1} + a_n $$
and $a_1=0=a_2$.
Now let us consider the kernel. This is the same thing, but with $b=(0, 0, \dots).$ In every entry we have the equation
$$ 0 = a_{n+2} - a_{n+1} - a_n. $$
Thus, we get $a_{n+2}=a_{n+1}+ a_{n}$. This means, $a$ is uniquely determined by $a_1$ and $a_2$ and hence, the kernel has dimension $2$. For example we could take $a_1=1=a_2$, then we get the Fibonacci numbers patched together. We could complete the vector with $a=(0,1,1,2, \dots)$, i.e. the previous vector shifted to the right and with an additional zero.
