Find the formula of Lucas numbers using the formula of Fibonacci numbers I have to find the formula for Lucas numbers using the formula for Fibonacci numbers.
Lucas numbers are numbers like Fibonacci numbers but $L_0=2$ and $L_1=1$;  other numbers are the sum of the previous two numbers. Now we know the Fibonacci numbers formula, but how to find the formula for Lucas numbers?
 A: Let we solve a more general problem.
Since both Lucas numbers and Fibonacci numbers obey
$$ A_{n+2} = A_{n+1}+A_{n} \tag{1}$$
the associated sequences have the same characteristic polynomial and
$$ A_n = k_1 \varphi^n + k_2 \bar{\varphi}^n $$
for some constants $k_1,k_2$, with $\varphi,\bar{\varphi}$ being the roots of $x^2-x-1$. It follows that the set of sequences fulfilling $(1)$ is a vector space with dimension $2$ and for every sequence obeying $(1)$ we have
$$ A_n = j_1 F_n + j_2 L_n \tag{2} $$
or, equivalently,
$$ A_n = \tau_1 F_n + \tau_2 F_{n+1},\qquad A_n = \eta_1 L_n + \eta_2 L_{n+1}.\tag{3}$$
Since $F_0=0, F_1=1, L_0=2, L_1=1$, we have:

$$ F_n = \frac{\varphi^n-\bar{\varphi}^n}{\sqrt{5}},\qquad L_n = \varphi^n+\bar{\varphi}^n, $$
$$ L_n = -F_n+2 F_{n+1},\qquad F_n = \frac{-L_n+2L_{n+1}}{5}. \tag{4} $$

Viète's theorem provides the key relations $\varphi+\bar{\varphi}=1$ and $\varphi\bar{\varphi}=-1$. Identities in $(4)$ (and many others) can also be derived from
$$ \begin{pmatrix}A_{n+2} \\ A_{n+1}\end{pmatrix}=\begin{pmatrix}1 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}A_{n+1} \\ A_{n}\end{pmatrix},\tag{5} $$
just by noticing that $x^2-x-1$ is also the characteristic polynomial of the involved matrix.
