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?
Best Answer
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 $n=1$: $$\begin{cases} X_0=A + B\\ X_1=A + 2 B \end{cases}\implies \begin{cases} B=X_1-X_0\\ A=X_0-B=2X_0-X_1 \end{cases}.$$