A strange equality of the operator E ($Eu_n=u_{n+1}$)

Question

The operator $E$ is defined as $Eu_n=u_{n+1}$.

I encountered a strange equality. when I tried out Let $u_n$ represent a series such that

$$u_{n+2}=u_{n+1}+u_n. \tag{$\star$}$$

Or $E^2{u_n}=Eu_{n+1}+Eu_n$

$\left(E^2-E-1\right)u_n=0$

$E\equiv \frac{1\pm \sqrt{5}}{2} $

What interpretation should be made on such an equality ?.

This equality seems interesting to me as two well know sequences that obeys$\huge\star$ Fibonacci sequence($u_1=u_2=1$) and Lucas sequence($u_1=1,u_2=3$) has it's $n$ th term represented by

$\Large {\frac{\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n}{\sqrt{5}}}$ and $\Large {\frac{\left(\frac{1+\sqrt{5}}{2}\right)^n+\left(\frac{1-\sqrt{5}}{2}\right)^n}{2}}$ respectively

If I am to follow that $E$ and $\frac{1\pm \sqrt{5}}{2} $. Then in case of the above two particular case of the recursion . The formula for the $n$ the term is not true(Of course I can't determine which sign($\pm$) should be taken and hence such an argument is bad).But interestingly , when we assume the same value for $E$, $\huge \star$ is valid with both the signs(the $\pm$).

I am really interested with such an equivalence, as I think it's interpretation can be exploited for a proof of those formulas. And also possibly a general solution. As

$E^n{u_n}=u_{n+1}$

Answer 1

$E$ acts on the finite-dimensional vector space consisting of sequences satisfying that recurrence relation, and those are its eigenvalues on that space. More generally, $E$ acts on the nullspace of $p(E)$ for any polynomial $p$ with eigenvalues the roots of $p$, and you can use this together with the theory of Jordan normal form to write down general solutions to linear recurrence relations.