For a simple Leslie Matrix and asociated system we can get the value of a vector $x_0$ by using $X_{n+1}=AX_N$. For example with the following system $X_{N+1}=AX_N=\begin{pmatrix} 0 & 3 & 5\\ 0.2 & 0 & 0\\ 0 & 0.4 & 0 \end{pmatrix}\begin{pmatrix} c_n\\ j_n\\ a_n \end{pmatrix}$ $\begin{cases} c_{n+1}=3j_n+5a_n\\ j_{n+1}=0.2c_n\\ a_{n+1}=0.4j_n \end{cases}$
But I noticed that by making some substitutions I get to a third order recurrence relations for $c_n$: $c_{n+3}=0.6c_{n+1}+0.4c_n$ and its characteristic equation is the same as the matrix one. (with roots $\lambda_1=1 $, $\lambda_2=-0.5+0.38i$, $\lambda_3=-0.5-0.38i$)
But how I can use only this recurrence to get a $\begin{pmatrix} c_n\\ j_n\\ a_n \end{pmatrix}$ vector from some $x_0$? as I do with the matrix, I think I´m strugling with complex numbers in development of solutions. For example with $x_0=\begin{pmatrix} 10\\ 8\\ 12 \end{pmatrix}$ by making $A^3X_0$ I get $x_3=$$\begin{pmatrix} 54.4\\ 4.4\\ 6.72 \end{pmatrix}$
How can I do the same without using the matrix, only with the $c_n$ recurrence relation?
$spectrum(A)=\{1,a,\overline{a}\}$ where $a=-1/2+i\sqrt{15}/10$. satisfies $|a|<1$. In this case, $A^n$ depends on the Im,Re parts of $a^n$. Then we have no closed forms for $A^n$ (in the sense that $(a+ib)^n+(a-ib)^n$ has no REAL closed form); if we want the EXACT value of $A^n$, then we must use some recurrence calculation.
Let $c=Re(a^n),f=Im(a^n)$. We obtain
where $c,f$ are given by the recurrence
Yet, if we want only an APPROXIMATION of $A^n$, then it suffices to ask for an approximation of $c,f$, that is much faster. It is easy to check the validity of the result, using the fact that $5^nA^n$ is an integer matrix.