Question 7.6.3: Consider the dynamical system x(t + 1) =[3 −5 1 −1] x(t) wit......

In Example 8 of Section 7.5, we found the eigenvalues

λ_{1,2} = 1 ± i.

The polar coordinates of eigenvalue 1 + i are r = \sqrt{2} and θ = \frac{π}{4} . Furthermore, we found that

S=\begin{bmatrix}\vec{w}&\vec{v}\end{bmatrix}=\begin{bmatrix}0&-5\\1&-2\end{bmatrix}.

Since

S^{-1}\vec{x}_0=\begin{bmatrix}1\\0\end{bmatrix},

Theorem 7.6.3 gives

\vec{x}(t)=(\sqrt{2})^t\begin{bmatrix}0&-5\\1&-2\end{bmatrix}\begin{bmatrix}\cos(\frac{\pi}{4}t)&-\sin(\frac{\pi}{4}t)\\ \sin(\frac{\pi}{4}t)&\cos(\frac{\pi}{4}t)\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}

=(\sqrt{2})^t\begin{bmatrix}-5\sin(\frac{\pi}{4}t)\\ \cos(\frac{\pi}{4}t)-2\sin(\frac{\pi}{4}t)\end{bmatrix}.

We leave it to the reader to work out the details of this computation.

Next, let’s think about the trajectory. We will develop the trajectory step by step:

• The points

\begin{bmatrix}\cos(\frac{\pi}{4}t)&-\sin(\frac{\pi}{4})\\ \sin(\frac{\pi}{4}t)&\cos(\frac{\pi}{4}t)\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}    (for t= 0, 1, 2, . . .)

are located on the unit circle, as shown in Figure 3a. Note that at t = 8 the system returns to its initial position, \begin{bmatrix}1\\0\end{bmatrix}; the period of this system is 8.

• In Exercise 2.2.54, we saw that an invertible linear transformation maps the unit circle into an ellipse. Thus, the points

\begin{bmatrix}0&-5\\1&-2\end{bmatrix}\begin{bmatrix}\cos(\frac{\pi}{4}t)&-\sin(\frac{\pi}{4}t)\\ \sin(\frac{\pi}{4}t)&\cos(\frac{\pi}{4}t)\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}

are located on an ellipse, as shown in Figure 3b. The two column vectors of

S=\begin{bmatrix}0&-5\\1&-2\end{bmatrix}=\begin{bmatrix}\vec{w}&\vec{v}\end{bmatrix}

are shown in that figure as well. Again, the period of this system is 8.

• The exponential growth factor (\sqrt{2})^t will produce longer and longer vectors

\vec{x}(t)=(\sqrt{2})^t\begin{bmatrix}0&-5\\1&-2\end{bmatrix}\begin{bmatrix}\cos(\frac{\pi}{4}t)&-\sin(\frac{\pi}{4}t)\\ \sin(\frac{\pi}{4}t)&\cos(\frac{\pi}{4}t)\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}.

Thus, the trajectory spirals outward, as shown in Figure 3c. (We are using different scales in Figures 3a, b, and c.) Note that \vec{x}(8) = (\sqrt{2})^8\vec{x}(0) = 16\vec{x}(0).