College Algebra: Find real-valued closed formulas for the trajectory $x(t+1)=Ax(t)$ ??

Question

Hey so I have this problem on my webwork that I do not understand:

The problem says to find real-valued closed formulas for the trajectory: $x(t+1)=Ax(t)$ where

$A=\begin{bmatrix} -0.8 & 0.6 \\ -0.6 & -0.8 \end{bmatrix}$ and $\overrightarrow{x}(0) = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$

My thinking is that I have to make the dynamical system $x_{k+1} = Ax_k$

I solved for the eigenvalues $\lambda_{1} = \frac{-4}{5} + \frac35i$ with its respective eigenvector $\begin{bmatrix} -i \\ 1 \end{bmatrix}$

and $\lambda_{2} = \frac{-4}{5} - \frac35i$ with its respective eigenvector $\begin{bmatrix} i \\ 1 \end{bmatrix}$

I made a formula $x_k = 1(\frac{-4}{5} + \frac35i)^k\begin{bmatrix} -i \\ 1 \end{bmatrix} + 0(\frac{-4}{5} - \frac35i)^k\begin{bmatrix} i \\ 1 \end{bmatrix}$ which evaluates to

$x_k = 1(\frac{-4}{5} + \frac35i)^k\begin{bmatrix} -i \\ 1 \end{bmatrix}$

but the answer is looking for a vector with 2 rows and 1 column.. what am I doing wrong??

Answer 1

Hint: $A$ is rotation matrix for angle $\theta = \pi + \arccos 0.8 \approx 3.785$ radians, which is approximate $216.9^{\circ}$. So $x(1)$ is $x(0)$, but rotated by this angle, $x(2)$ is $x(0)$ rotated by $2\theta$ and so on. You can write in as \begin{align} x(k) = A^k x(0) &= \begin{pmatrix} \cos{k(\pi + \arccos 0.8)} & -\sin{k(\pi + \arccos 0.8)}\\ \sin{k(\pi + \arccos 0.8)} & \cos{k(\pi + \arccos 0.8)} \end{pmatrix} x(0) \\&= \begin{pmatrix} (-1)^k cos{(k\arccos 0.8)} & - (-1)^k\sin{(k\arccos 0.8)} \\ (-1)^k\sin{(k\arccos 0.8)} & (-1)^k \cos{(k\arccos 0.8)} \end{pmatrix} x(0) \\&= (-1)^{k} \begin{pmatrix} \cos{(k\arccos 0.8)} & -\sin{(k\arccos 0.8)} \\ \sin{(k\arccos 0.8)} & \cos{(k\arccos 0.8)} \end{pmatrix} x(0) \end{align}