linear recursion $y_n=A \cdot y_{n-1}$

Question

Let $a,b, \in \mathbb{R}$. Let $x_0=a, x_1=b$ and $x_n=\frac{x_{n-1}+x_{n-2}}{2}$ for $n \geq 2$

(i) Write the recursion in the form $y_n=A \cdot y_{n-1}$ where $A$ is a $2 \times 2$ matrix and $y_i=\begin {pmatrix} x_i \\ x_{i-1} \end{pmatrix}$

(ii) Find a diagonal matrix $D$ and an invertible matrix $S$, so that $A=SDS^{-1}$

(iii) Calculate $\lim_{n \rightarrow \infty}S^{-1}A^nS$

(iv) Deduce from that $\lim_{n \rightarrow \infty}A^n$ and $\lim_{n \rightarrow \infty}x_n$

My solution so far:

(i) $\begin {pmatrix} x_i \\ x_{i-1} \end{pmatrix}=\begin {pmatrix} \frac{1}{2} & \frac{1}{2} \\ 1 & 0 \end{pmatrix} \begin {pmatrix} x_{i-1} \\ x_{i-2} \end{pmatrix}$

(ii) Eigenvalues are $\lambda_1=1$ and $\lambda_1=-\frac{1}{2}$.

Basis of $V_{\lambda_1}$ is $B_{\lambda_1} =\begin {pmatrix} 1 \\ 1 \end{pmatrix}$

Basis of $V_{\lambda_2}$ is $B_{\lambda_2} =\begin {pmatrix} -1 \\ 2 \end{pmatrix}$

$\Rightarrow D=\begin {pmatrix} 1 & 0 \\ 0 & -\frac{1}{2} \end{pmatrix}, S\begin {pmatrix} 1 & -1 \\ 1 & 2 \end{pmatrix}, S^{-1}= \begin {pmatrix} \frac{2}{3} & \frac{1}{3} \\ -\frac{1}{3} & \frac{1}{3} \end{pmatrix}$

(iii) $\lim_{n \rightarrow \infty}S^{-1}A^nS = \lim_{n \rightarrow \infty}S^{-1}(SDS^{-1})^nS=\lim_{n \rightarrow \infty}S^{-1}(SD^nS^{-1})S=\lim_{n \rightarrow \infty}D^n= \lim_{n \rightarrow \infty} \begin {pmatrix} 1 & 0 \\ 0 & (-\frac{1}{2})^n \end{pmatrix}$

Is this correct so far? How to go on now?

Answer 1

Because $\lim_{n \rightarrow \infty}(-\frac{1}{2})^n=0$, we have

$$\lim_{n \rightarrow \infty}D^n= \lim_{n \rightarrow \infty} \begin {pmatrix} 1 & 0 \\ 0 & (-\frac{1}{2})^n \end{pmatrix}=\begin {pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$$

To simplify the notation, we Define $X^\infty:=\lim_{n \rightarrow \infty}X^n$ and $Y_\infty:=\lim_{n \rightarrow \infty}Y_n$

Then $$A^\infty=SD^\infty S^{-1}=\begin {pmatrix} \frac{2}{3} & \frac{1}{3} \\ \frac{2}{3} & \frac{1}{3} \end{pmatrix}$$

Therefore (with $x_1=\{a,b\}$) $$y_\infty=A^\infty.x_1$$

From any component of vector $y_\infty$, we have $$x_\infty=\frac{2}{3}a+\frac{1}{3}b$$

This is consistent because $\lim_{n \rightarrow \infty}x_n=\lim_{n \rightarrow \infty}x_{n-1}$.