# Recursive formula, eigenvalue problem

A sequence $$\{X_i \}_{i \geq 0}$$ is defined recursive by $$X_{n+1} = 3 X_{n} - 2X_{n-1}, \qquad n \geq1.$$

and $$X_0 = \begin{pmatrix} 1 & 0 \newline 1 & 1 \end{pmatrix},X_1 = \begin{pmatrix} 1 & 1 \newline 1 & 1 \end{pmatrix}.$$ Find an explicit formula for $$X_n$$.

My attempt;

We can write $$= \begin{pmatrix} X_{n+1} \newline X_{n} \end{pmatrix} = \begin{pmatrix} 3 & -2 \newline 1 & 0 \end{pmatrix} \begin{pmatrix} X_{n} \newline X_{n-1} \end{pmatrix}$$ for $$n \geq 1.$$ Diagonalizing the matrix yields

$$\begin{pmatrix} X_{n+1} \newline X_{n} \end{pmatrix} = \begin{pmatrix} 2 & 1 \newline 1 & 1 \end{pmatrix} \begin{pmatrix} 2^n & 0 \newline 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 \newline -1 & 2 \end{pmatrix} \begin{pmatrix} X_{n} \newline X_{n-1} \end{pmatrix}$$

This is and old exam; now the answer concludes that $$X_n$$ can be written as $$X_n = A + 2^n B$$, for some matrices A and B, which can be found by plugging in $$X_1$$ and $$X_0.$$ My question is how did he draw this conclusion, is it only from the eigenvalues? How would one approach questions on this form if the matrix couldnt be diagonalized?

We may prove the claim $$X_n = A + 2^n B$$ by induction. The inductive step is easy: for $$n\geq 1$$, $$X_{n+1} = 3 X_{n} - 2X_{n-1}=3(A + 2^n B)-2(A + 2^{n-1} B)=A + 2^{n+1} B.$$ It remains to find $$A$$ and $$B$$ by considering the cases $$n=0$$ and $$n=1$$: $$\begin{cases} X_0=A + B\\ X_1=A + 2 B \end{cases}\implies \begin{cases} B=X_1-X_0\\ A=X_0-B=2X_0-X_1 \end{cases}.$$
• Where does $(A + 2^n B)$ come from and why not $2X_{n-1} = 2(A+2^{n-1} B)?$ Apr 8, 2020 at 18:53
• As regards the first question. For a linear recursion $X_{n+1}=a X_{n} +bX_{n-1}$ the roots of the polynomial $z^2-az-b=0$ play an important role. In your case the roots are $1$ and $2$. See en.wikipedia.org/wiki/… Apr 8, 2020 at 18:58
Note that $$X_{n+1}-2 X_n = X_n - 2 X_{n-1} = \Delta = X_1 - 2 X_0$$.
Hence $$X_{n+1} = 2 X_n + \Delta$$.
The general solution is $$X_n = 2^n X_0 + \sum_{k=0}^{n-1} 2^{n-k-1} \Delta = 2^n X_0+(2^{n}-1)\Delta = 2^n(X_1-X_0)-X_1+2X_0$$.