Compute the solution $\phi_t \overrightarrow x_0 = e^{At} \overrightarrow x_0$ to the system $x' = -x + 4y$ and $y' = -4x - y$ Compute the solution $\phi_t \overrightarrow x_0 = e^{At} \overrightarrow x_0$ to the system $x' = -x + 4y$ and $y' = -4x - y$
The characteristic equation i found is $\lambda^2 + 2\lambda + 17$...The eienvalues are -1+- 4i....can anyone help me to move forward since i do not know how to find the eigenvector from non real eigenvalues?
 A: The eigenvalues are:
$$\lambda_{1,2} = -1 ~ \pm ~ 4i$$
To find the eigenvectors, we solve:
$$[A - \lambda I]v_i = 0$$
For the first eigenvalue, we have:
$$\begin{bmatrix}-1 -(-1 + 4i) & 4\\-4 & -1 -(-1 + 4i)\end{bmatrix}v_1 = 0 \implies \begin{bmatrix}- 4i & 4\\-4 & - 4i\end{bmatrix}v_1 = 0$$
Finding the RREF $(R_2 = R_2 + iR_1, R_1 = R_1/4, R_1 = R_1/i)$ of this matrix yields:
$$\begin{bmatrix}1 & i\\0 & 0\end{bmatrix}v_1 = 0$$
This gives us the eigenvector:
$$v_1 = (-i, 1)$$
Since we have a complex conjugate eigenvalue, the other eigenvector is:
$$v_2 = (i, 1)$$
There are many ways to find $e^{A t}$. We can do:
$$e^{A t} = P e^{D t}P^{-1} = \begin{bmatrix}-i & i\\1 & 1\end{bmatrix}~\begin{bmatrix}e^{(-1 + 4i)t} & 0\\0 & e^{(-1 - 4i)t}\end{bmatrix}~\begin{bmatrix}\dfrac{i}{2} & \dfrac{1}{2}\\-\dfrac{i}{2} & \dfrac{1}{2}\end{bmatrix} = e^{-t}\begin{bmatrix}\cos 4t & \sin 4t\\-\sin 4t & \cos 4t\end{bmatrix}$$
We could have also found a fundamental matrix $M(t)$ and formed $e^{At} = M(t) M^{-1}(0)$ and other approaches.
Update
If we plot the phase portrait for this system, we have:

What do you notice about any initial condition? What can you say about the stability?
