We now consider the system \({\bf y}'=A{\bf y}\), where \(A\) has a complex eigenvalue \(\lambda=\alpha+i\beta\) with \(\beta\ne0\). We continue to assume that \(A\) has real entries, so the characteristic polynomial of \(A\) has real coefficients. This implies that \(\overline\lambda=\alpha-i\beta\) is also an eigenvalue of \(A\).
An eigenvector \({\bf x}\) of \(A\) associated with \(\lambda=\alpha+i\beta\) will have complex entries, so we’ll write
\[{\bf x}={\bf u}+i{\bf v} \nonumber \]
where \({\bf u}\) and \({\bf v}\) have real entries; that is, \({\bf u}\) and \({\bf v}\) are the real and imaginary parts of \({\bf x}\). Since \(A{\bf x}=\lambda {\bf x}\),
which shows that \({\bf x}={\bf u}-i{\bf v}\) is an eigenvector associated with \(\overline\lambda=\alpha-i\beta\). The complex conjugate eigenvalues \(\lambda\) and \(\overline\lambda\) can be separately associated with linearly independent solutions \({\bf y}'=A{\bf y}\); however, we will not pursue this approach, since solutions obtained in this way turn out to be complex–valued. Instead, we’ll obtain solutions of \({\bf y}'=A{\bf y}\) in the form
where \(f_1\) and \(f_2\) are real–valued scalar functions. The next theorem shows how to do this.
Theorem 10.6.1
Let \(A\) be an \(n\times n\) matrix with real entries\(.\) Let \(\lambda=\alpha+i\beta\) (\(\beta\ne0\)) be a complex eigenvalue of \(A\) and let \({\bf x}={\bf u}+i{\bf v}\) be an associated eigenvector\(,\) where \({\bf u}\) and \({\bf v}\) have real components\(.\) Then \({\bf u}\) and \({\bf v}\) are both nonzero and
Carrying out the multiplication indicated on the right side of Equation \ref{eq:10.6.1} and collecting the real and imaginary parts of the result yields
We leave it to you (Exercise 10.6.25) to show from this that \({\bf u}\) and \({\bf v}\) are both nonzero. Substituting from these equations into Equation \ref{eq:10.6.4} yields
is a solution of \({\bf y}'=A{\bf y}\) for any choice of the constants \(c_1\) and \(c_2\). In particular, by first taking \(c_1=1\) and \(c_2=0\) and then taking \(c_1=0\) and \(c_2=1\), we see that \({\bf y}_1\) and \({\bf y}_2\) are solutions of \({\bf y}'=A{\bf y}\). We leave it to you to verify that they are, respectively, the real and imaginary parts of Equation \ref{eq:10.6.3} (Exercise 10.6.26), and that they are linearly independent (Exercise 10.6.27).
Hence, \(\lambda=1+4i\) is an eigenvalue of \(A\). The associated eigenvectors satisfy \(\left(A-\left(1+4i\right)I\right){\bf x}={\bf 0}\). The augmented matrix of this system is
Hence, \(\lambda=1+3i\) is an eigenvalue of \(A\). The associated eigenvectors satisfy \(\left(A-(1+3i)I\right){\bf x}={\bf 0}\). The augmented augmented matrix of this system is
\(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) is a fundamental set of solutions of Equation \ref{eq:10.6.8}. The general solution of Equation \ref{eq:10.6.8} is
\(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) is a fundamental set of solutions of Equation \ref{eq:10.6.9}. The general solution of Equation \ref{eq:10.6.9} is
has a complex eigenvalue \(\lambda=\alpha+i\beta\) (\(\beta\ne0\)) and \({\bf x}={\bf u}+i{\bf v}\) is an associated eigenvector, where \({\bf u}\) and \({\bf v}\) have real components. To describe the trajectories accurately it is necessary to introduce a new rectangular coordinate system in the \(y_1\)-\(y_2\) plane. This raises a point that hasn’t come up before: It is always possible to choose \({\bf x}\) so that \(({\bf u},{\bf v})=0\). A special effort is required to do this, since not every eigenvector has this property. However, if we know an eigenvector that doesn’t, we can multiply it by a suitable complex constant to obtain one that does. To see this, note that if \({\bf x}\) is a \(\lambda\)-eigenvector of \(A\) and \(k\) is an arbitrary real number, then
If \(({\bf u},{\bf v})\ne0\) we can use the quadratic formula to find two real values of \(k\) such that \(({\bf u}_1,{\bf v}_1)=0\) (Exercise 10.6.28).
for the matrix of the system Equation \ref{eq:10.6.6}. Here \(\bf {u}=\twocol{3}{5}\) and \({\bf v}=\twocol40\) are not orthogonal, since \(({\bf u},{\bf v})=12\). Since \(\|{\bf v}\|^2-\|{\bf u}\|^2=-18\), Equation \ref{eq:10.6.12} is equivalent to
\[2k^2-3k-2=0. \nonumber \]
The zeros of this equation are \(k_1=2\) and \(k_2=-1/2\). Letting \(k=2\) in Equation \ref{eq:10.6.11} yields
(The numbers don’t always work out as nicely as in this example. You’ll need a calculator or computer to do Exercises 10.6.29-10.6.40.) Henceforth, we’ll assume that \(({\bf u},{\bf v})=0\). Let \({\bf U}\) and \({\bf V}\) be unit vectors in the directions of \({\bf u}\) and \({\bf v}\), respectively; that is, \({\bf U}={\bf u}/\|{\bf u}\|\) and \({\bf V}={\bf v}/\|{\bf v}\|\). The new rectangular coordinate system will have the same origin as the \(y_1\)-\(y_2\) system. The coordinates of a point in this system will be denoted by \((z_1,z_2)\), where \(z_1\) and \(z_2\) are the displacements in the directions of \({\bf U}\) and \({\bf V}\), respectively. From Equation \ref{eq:10.6.5}, the solutions of Equation \ref{eq:10.6.10} are given by
For convenience, let’s call the curve traversed by \(e^{-\alpha t}{\bf y}(t)\) a shadow trajectory of Equation \ref{eq:10.6.10}. Multiplying Equation \ref{eq:10.6.13} by \(e^{-\alpha t}\) yields
which means that the shadow trajectories of Equation \ref{eq:10.6.10} are ellipses centered at the origin, with axes of symmetry parallel to \({\bf U}\) and \({\bf V}\). Since
the vector from the origin to a point on the shadow ellipse rotates in the same direction that \({\bf V}\) would have to be rotated by \(\pi/2\) radians to bring it into coincidence with \({\bf U}\) (Figures 10.6.1
and 10.6.2
).
If \(\alpha=0\), then any trajectory of Equation \ref{eq:10.6.10} is a shadow trajectory of Equation \ref{eq:10.6.10}; therefore, if \(\lambda\) is purely imaginary, then the trajectories of Equation \ref{eq:10.6.10} are ellipses traversed periodically as indicated in Figures 10.6.1
and 10.6.2
. If \(\alpha>0\), then
so the trajectory spirals away from the origin as \(t\) varies from \(-\infty\) to \(\infty\). The direction of the spiral depends upon the relative orientation of \({\bf U}\) and \({\bf V}\), as shown in Figures 10.6.3
and 10.6.4
. If \(\alpha<0\), then
so the trajectory spirals toward the origin as \(t\) varies from \(-\infty\) to \(\infty\). Again, the direction of the spiral depends upon the relative orientation of \({\bf U}\) and \({\bf V}\), as shown in Figures 10.6.5
and 10.6.6
.