Suppose we take the linear differential equation
\begin{align*}
x^\prime (t) \amp = 2x (t) + y(t), \\
y^\prime (t) \amp = -2 x (t) + 4 y(t).
\end{align*}
Writing this out we find the matrix
\begin{equation*}
A = \left[ \begin{matrix} 2 \amp 1 \\ -2 \amp 4 \end{matrix} \right]
\end{equation*}
and obtain the equation
\begin{equation*}
\mb{x}^\prime (t) = A \mb{x} (t).
\end{equation*}
Again, the right hand side is the vector field
\begin{equation*}
\mb{F} (\mb{x} ) = A \mb{x}
\end{equation*}
and it would be nice to diagonalize \(A\) to get a picture of this field. However, computing the characteristic polynomial gives
\begin{equation*}
p_A (s) = \det \left[ \begin{matrix} s -2 \amp -1 \\ 2 \amp s - 4 \end{matrix} \right] = s^2 - 6 s + 10.
\end{equation*}
One can use the quadratic formula here and see that
\begin{equation*}
p_A (s) = (s - (3 + i)) (s - (3 - i))
\end{equation*}
so that the roots of \(p_A (s)\) are complex numbers and \(A\) cannot be diagonalized as a real matrix. Now, one of the benefits of having real eigenvalues and eigenvectors was that we obtained a nice picture of the vector field. There is something to be said in the case of a complex eigenvector as well, which is related to several of the exercises that you have worked through. In particular, if there is an eigenvalue of the form \(\lambda = re^{i\theta}\text{,}\) then one will see a \(\theta\) rotation in the vectors of the vector field (in some coordinate system). For now, we will leave the illustration of this field to a computer and come back to the geometry once we’ve found the solution.
To compute the general solution, there are several ways to proceed, but we will take a principled approach and simply say that our path \(\mb{x}\) was a function to \(\mathbb{C}^2\) all along so that
\begin{equation*}
\mb{x} : I \to \mathbb{C}^2.
\end{equation*}
Now we can diagonalize because the eigenvalues are distinct. A bit of computation gives the eigenbasis
\begin{equation*}
\mathcal{B} = \left\{ \mb{v}_1 , \mb{v}_2 \right\} = \left\{ \twovec{i}{-1 + i}, \twovec{-i}{-1 - i} \right\}.
\end{equation*}
Again, we can write our solution in terms of this basis
\begin{align}
\mb{x} \amp = \twovec{\tilde{x} (t)}{\tilde{y} (t)}_\mathcal{B} \tag{5.3.4}\\
\amp = \tilde{x} (t) \mb{v}_1 + \tilde{y} (t) \mb{v}_2 \tag{5.3.5}\\
\amp = \twovec{i \tilde{x} (t) - i \tilde{y} (t)}{(-1 + i) \tilde{x} (t) + (-1 - i) \tilde{y} (t)} \tag{5.3.6}
\end{align}
and again, in this basis the differential equation becomes much simpler
\begin{equation*}
\twovec{\tilde{x}^\prime (t)}{\tilde{y}^\prime (t)} = \twovec{ (3 + i) \tilde{x} (t)}{(3 - i) \tilde{y} (t)}_{\mathcal{B}} .
\end{equation*}
This time, the two independent equations are
\begin{align*}
\tilde{x}^\prime (t) \amp = (3 + i) \tilde{x} (t), \\
\tilde{y}^\prime (t) \amp = (3 - i) \tilde{y} (t).
\end{align*}
It is here that
Exercise 5.1.4.3 comes into view and we realize that, in fact, we’ve already solved these equations and obtained
\begin{align*}
\tilde{x}(t) \amp = C_1 e^{(3 + i)t}, \\
\tilde{y} (t) \amp = C_2 e^{(3 - i)t}.
\end{align*}
So, were we inclined to write our solutions in terms of the basis \(\mathcal{B}\) in \(\mathbb{C}^2\text{,}\) we would have solved the differential equation with
\begin{gather*}
\mb{x} (t) = \twovec{ C_1 e^{(3 + i)t}}{C_2 e^{(3 - i)t}}_{\mathcal{B}} \text{ with initial condition } \mb{x} (0) = \twovec{C_1}{C_2}_{\mathcal{B}}.
\end{gather*}
I can hear the overwhelming chorus of objections to this solution from engineer and mathematician alike. After all, the constants \(C_1\) and \(C_2\) are possibly complex numbers and the function is too. We started in the real plane and have ended in what appears to be a terrifying \(4\)-dimensional mess (since \(\mathbb{C}^2\) is \(4\) real dimensions). Well, I contend that appearances can be deceiving! Let’s unwind this a bit with a few simple observations.
The first thing to recognize about our solution is that the basis \(\mathcal{B}\) which we chose had a hidden symmetry. Namely, if we take the complex conjugate of \(\mb{v}_1\) we obtain \(\mb{v}_2\) so that
\begin{equation*}
\bar{\mb{v}}_1 = \mb{v}_2.
\end{equation*}
Now, the solution we obtain for \(\mb{x}\) will be real, which means it equals its conjugate. Thus
\begin{align*}
\tilde{x} (t) \mb{v}_1 + \tilde{y} (t) \mb{v}_2 \amp = \overline{\tilde{x} (t) \mb{v}_1 + \tilde{y} (t) \mb{v}_2}, \\
\amp = \overline{\tilde{x} (t)} \bar{\mb{v}}_1 + \overline{\tilde{y} (t)} \bar{\mb{v}}_2,\\
\amp = \overline{\tilde{y} (t)} \mb{v}_1 + \overline{\tilde{x} (t)} \mb{v}_2,
\end{align*}
But since our coefficients are unique, we have that
\begin{equation}
\overline{\tilde{x} (t)} = \tilde{y} (t).\tag{5.3.7}
\end{equation}
Putting our solution in to this equation gives
\begin{equation*}
\bar{C}_1 = C_2.
\end{equation*}
Even more helpful, we note that for any complex number \(z\) we can get the real and imaginary parts of \(z\) by simply checking that
\begin{align*}
\operatorname{Re} (z) \amp = \frac{1}{2} \left( z + \bar{z} \right), \\
\operatorname{Im} (z) \amp = - \frac{i}{2} \left( z - \bar{z} \right).
\end{align*}
\begin{equation*}
\mb{x} (t) = \twovec{ -2 \operatorname{Im} (\tilde{x} (t))}{-2 \operatorname{Re} (\tilde{x} (t)) - 2\operatorname{Im} (\tilde{x} (t)) } = \twovec{ \operatorname{Im} ( -2\tilde{x} (t))}{ \operatorname{Re} (-2\tilde{x} (t)) + \operatorname{Im} (-2\tilde{x} (t)) }.
\end{equation*}
Now, note that \(C_1\) is zero only for the zero solution, so assuming it is not zero, we can find an \(a, b\) so that in polar coordinates
\begin{equation*}
-2C_1 = e^{a + bi}.
\end{equation*}
This gives us a nice way to rewrite our solution as
\begin{equation*}
-2\tilde{x} (t) = -2 C_1 e^{(3 + i)t} = e^{(3t + a) + (t + b)i} = e^{3t + a} \cos (t + b) + i e^{3t + a} \sin (t + b).
\end{equation*}
Pulling out the real and imaginary parts, we obtain a very real looking solution
\begin{equation*}
\mb{x} (t) = \twovec{e^{3t + a} \sin (t + b)}{ e^{3t + a} \cos (t + b) + e^{3t + a} \sin (t + b)}.
\end{equation*}
To finish the solution, one would have to solve \(a\) and \(b\) for an initial condition
\begin{equation*}
\mb{x} (0) = \twovec{A}{B}
\end{equation*}
but this will be left as a linear algebra exercise for the student (which can be done in multiple ways).
However, it is interesting to note here that the solution \(\mb{x} (t)\) can be written
\begin{equation*}
\mb{x} (t) = e^{3t + a} \twovec{ \sin (t + b)}{ \cos (t + b) + \sin (t + b)}.
\end{equation*}
The scaling factor increases exponentially which means that the solution will head off to infinity. What about the vector portion? Well, it is not hard to see that this is a parameterization of the conic section
\begin{equation*}
2x^2 - 2xy + y^2 = 1.
\end{equation*}
A meticulous student will check and see that this equation is that of a (rotated) ellipse. So the solution is simply following a parameterization of an ellipse, but scaling it simultaneously and spiraling away from the origin.
