Assume $E$ is the difference equation. We have that $E^3(E-2)=0$
The solution should be $x=(a_1\ a_2\ a_3\ 2a_4\ 4a_4 … 2^n a_4…)$ But what is the basis of the solution space?
I think it should be $(0\ 0\ 0\ 2\ 4\ 8 …2^n…)$. Because $a_1,a_2,a_3$ are solutions to be fixed.(i.e. initial solution)
But it appears another possibility that $$(1\ 0\ 0\ …)$$ $$(0\ 1\ 0\ …)$$ $$(0\ 0\ 1\ …)$$ $$(0\ 0\ 0\ 2\ 4\ 8 …2^n…)$$ are the basis.
What do you guys think?
The (E-2) has as in its one-dimensional nullspace $v=(1,2,4,\ldots,2^n,\ldots)$, but the nullspace associated with $E^3$ is somewhat complicated because of the repeated eigenvalue of $0$; hence it should a dimension three nullspace, namely all sequences $(a,b,c,0,0,0,\ldots)$ for any $a,b,c$.
And so, a basis for the solution space is \begin{align*} &(1,0,0,0,\ldots)\\ &(0,1,0,0,\ldots)\\ &(0,0,1,0,\ldots)\\ &(1,2,4,\ldots,2^{n},\ldots) \end{align*}
Now, $\frac{1}{8}(v-(1,2,4,0,0,0\ldots))=(0,0,0,1,2,4,\ldots,2^{n-3},\ldots)$, so we can also write as a basis: \begin{align*} &(1,0,0,0,\ldots)\\ &(0,1,0,0,\ldots)\\ &(0,0,1,0,\ldots)\\ &(0,0,0,1,2,4,\ldots,2^{n-3},\ldots) \end{align*} which is almost as you wrote. But, we can scale any of these vectors as we wish (namely, divide the final vector by two).
After this, it's up to the initial conditions. Since this is a four-dimensional solution space, one would need four values from the sequence to define a particular solution. In the second set of basis vectors, the first four elements of the sequence correspond to the coefficients to these vectors.
