
# 4.6: Constant Coefficient Homogeneous Systems III

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## Constant Coefficient Homogeneous Systems III

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

\begin{eqnarray*}
{\bf x} = {\bf u} + i{\bf v}
\end{eqnarray*}

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 $$A{\bf x} = \lambda {\bf x}$$,

\label{eq:4.6.1}
A({\bf u}+i{\bf v})=(\alpha+i\beta)({\bf u}+i{\bf v}).

Taking complex conjugates here and recalling that $$A$$ has real entries yields

\begin{eqnarray*}
A ({\bf u} - i{\bf v}) = (\alpha - i\beta) ({\bf u} - i{\bf v}),
\end{eqnarray*}

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 won't 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

\label{eq:4.6.2}
{\bf y}=f_1{\bf u}+f_2{\bf v}

where $$f_1$$ and $$f_2$$ are real-valued scalar functions. The next theorem shows how to do this.

### Theorem $$\PageIndex{1}$$

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

\begin{eqnarray*}
{\bf y}_1 = e^{\alpha t} ({\bf u} \cos \beta t - {\bf v} \sin \beta t) \quad \mbox{and} \quad {\bf y}_2 = e^{\alpha t}({\bf u} \sin \beta t + {\bf v} \cos \beta t),
\end{eqnarray*}

which are the real and imaginary parts of

\label{eq:4.6.3}
e^{\alpha t}(\cos\beta t+i\sin\beta t)({\bf u}+i{\bf v}),

are linearly independent solutions of $${\bf y}'=A{\bf y}$$.

Proof

A function of the form \eqref{eq:4.6.2} is a solution of $${\bf y}'=A{\bf y}$$ if and only if

\label{eq:4.6.4}
f_1'{\bf u}+f_2'{\bf
v}=f_1A{\bf u}+f_2A{\bf v}.

Carrying out the multiplication indicated on the right side of \eqref{eq:4.6.1} and collecting the real and imaginary parts of the result yields

\begin{eqnarray*}
A ({\bf u} + i{\bf v}) = (\alpha {\bf u} - \beta {\bf v}) + i(\alpha {\bf v} + \beta {\bf u}).
\end{eqnarray*}

Equating real and imaginary parts on the two sides of this equation yields

\begin{eqnarray*}
\begin{array} \\ A {\bf u} &=& \alpha {\bf u} - \beta {\bf v} \\ A {\bf v} &=& \alpha {\bf v} + \beta {\bf u} \end{array}.
\end{eqnarray*}

We leave it to you (Exercise $$(4.6E.25)$$) to show from this that $${\bf u}$$ and $${\bf v}$$ are both nonzero. Substituting from these equations into \eqref{eq:4.6.4} yields

\begin{eqnarray*}
f_1'{\bf u}+f_2'{\bf v}
&=&f_1(\alpha{\bf u}-\beta{\bf v})+f_2(\alpha{\bf v}+\beta{\bf u})\\
&=&(\alpha f_1+\beta f_2){\bf u}+(-\beta f_1+\alpha f_2){\bf v}.
\end{eqnarray*}

This is true if

\begin{eqnarray*}
\begin{array} \\ f_1' &=& \alpha f_1 + \beta f_2 \phantom{,} \\ f_2' &=& -\beta f_1 + \alpha f_2, \end{array} \quad \mbox{or, equivalently,} \quad \begin{array} f_1' -\alpha f_1 &=& \phantom{-} \beta f_2 \phantom{.} \\ f_2' - \alpha f_2 &=& -\beta f_1. \end{array}
\end{eqnarray*}

If we let $$f_1=g_1e^{\alpha t}$$ and $$f_2=g_2e^{\alpha t}$$, where $$g_1$$ and $$g_2$$ are to be determined, then the last two equations become

\begin{eqnarray*}
\begin{array} \\ g_1' &=& \beta g_2 \phantom{.} \\ g_2' &=& -\beta g_1, \end{array}
\end{eqnarray*}

which implies that

\begin{eqnarray*}
g_1'' = \beta g_2' = -\beta^2 g_1,
\end{eqnarray*}

so

\begin{eqnarray*}
g_1'' + \beta^2 g_1 = 0
\end{eqnarray*}

The general solution of this equation is

\begin{eqnarray*}
g_1 = c_1 \cos \beta t + c_2 \sin \beta t.
\end{eqnarray*}

Moreover, since $$g_2=g_1'/\beta$$,

\begin{eqnarray*}
g_2 = -c_1 \sin \beta t + c_2 \cos \beta t.
\end{eqnarray*}

Multiplying $$g_1$$ and $$g_2$$ by $$e^{\alpha t}$$ shows that

\begin{eqnarray*}
f_1&=&e^{\alpha t}(\phantom{-}c_1\cos\beta t+c_2\sin\beta t ),\\
f_2&=&e^{\alpha t}(-c_1\sin\beta t+c_2\cos\beta t).
\end{eqnarray*}

Substituting these into \eqref{eq:4.6.2} shows that

\label{eq:4.6.5}
\begin{array}{rcl}
{\bf y}&=&e^{\alpha t}\left[(c_1\cos\beta t+c_2\sin\beta t){\bf u}
+(-c_1\sin\beta t+c_2\cos\beta t){\bf v}\right]\\
&=&c_1e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t)
+c_2e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t)
\end{array}

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 \eqref{eq:4.6.3} (Exercise $$(4.6E.26)$$), and that they are linearly independent (Exercise $$(4.6E.27)$$).

### Example $$\PageIndex{1}$$

Find the general solution of

\label{eq:4.6.6}
{\bf y}' = \left[ \begin{array} \\ 4 & {-5} \\ 5 & {-2} \end{array} \right] \bf y.

The characteristic polynomial of the coefficient matrix $$A$$ in \eqref{eq:4.6.6} is

\begin{eqnarray*}
\left| \begin{array} \\ 4 -\lambda & -5 \\ 5 & -2 - \lambda \end{array} \right| = (\lambda -1)^2 + 16.
\end{eqnarray*}

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

\begin{eqnarray*}
\left[ \begin{array} \\ 3-4i & -5 & \vdots & 0 \\ 5 & -3-4i & \vdots & 0 \end{array} \right],
\end{eqnarray*}

which is row equivalent to

\begin{eqnarray*}
\left[ \begin{array} \\ 1 & -{3+4i \over 5} & \vdots & 0 \\ 0 & 0 & \vdots & 0 \end{array} \right].
\end{eqnarray*}

Therefore $$x_1=(3+4i)x_2/5$$. Taking $$x_2=5$$ yields $$x_1=3+4i$$, so

\begin{eqnarray*}
{\bf x} = \left[ \begin{array} \\ 3+4i \\ 5 \end{array} \right]
\end{eqnarray*}

is an eigenvector. The real and imaginary parts of

\begin{eqnarray*}
e^t (\cos 4t + i \sin 4t) \left[ \begin{array} \\ 3+4i \\ 5 \end{array} \right]
\end{eqnarray*}

are

\begin{eqnarray*}
{\bf y}_1 = e^t \left[ \begin{array} \\ 3 \cos 4t-4 \sin 4t \\ 5 \cos 4t \end{array} \right] \quad \mbox{and} \quad {\bf y}_2 = e^t \left[ \begin{array} \\ 3 \sin 4t + 4 \cos 4t \\ 5 \sin 4t \end{array} \right],
\end{eqnarray*}

which are linearly independent solutions of \eqref{eq:4.6.6}. The general solution of \eqref{eq:4.6.6} is

\begin{eqnarray*}
{\bf y} = c_1 e^t \left[ \begin{array} \\ 3 \cos 4t-4 \sin 4t \\ 5 \cos 4t \end{array} \right] + c_2 e^t \left[ \begin{array} \\ 3 \sin 4t+4 \cos 4t \\ 5 \sin 4t \end{array} \right].
\end{eqnarray*}

### Example $$\PageIndex{2}$$

Find the general solution of

\label{eq:4.6.7}
{\bf y}' = \left[ \begin{array} \\ {-14} & {39} \\ {-6} & {16} \end{array} \right]{\bf y}.

The characteristic polynomial of the coefficient matrix $$A$$ in \eqref{eq:4.6.7} is

\begin{eqnarray*}
\left| \begin{array} \\ -14 - \lambda & 39 \\ -6 & 16 - \lambda \end{array} \right| = (\lambda - 1)^2 + 9.
\end{eqnarray*}

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

\begin{eqnarray*}
\left[ \begin{array} \\ -15-3i & 39 & \vdots & 0 \\ -6 & 15-3i & \vdots & 0 \end{array} \right],
\end{eqnarray*}

which is row equivalent to

\begin{eqnarray*}
\left[ \begin{array} \\ 1 & {-5+i \over 2} & \vdots & 0 \\ 0 & 0 & \vdots & 0 \end{array} \right].
\end{eqnarray*}

Therefore $$x_1=(5-i)/2$$. Taking $$x_2=2$$ yields $$x_1=5-i$$, so

\begin{eqnarray*}
{\bf x} = \left[ \begin{array} \\ 5-i \\ 2 \end{array} \right]
\end{eqnarray*}

is an eigenvector. The real and imaginary parts of

\begin{eqnarray*}
e^t (\cos 3t + i \sin 3t) \left[ \begin{array} \\ 5-i \\ 2 \end{array} \right]
\end{eqnarray*}

are

\begin{eqnarray*}
{\bf y}_1 = e^t \left[ \begin{array} \\ \sin 3t + 5 \cos 3t \\ 2 \cos 3t \end{array} \right] \quad \mbox{and} \quad {\bf y}_2 = e^t \left[ \begin{array} \\ -\cos 3t + 5 \sin 3t \\ 2 \sin 3t \end{array} \right],
\end{eqnarray*}

which are linearly independent solutions of \eqref{eq:4.6.7}. The general solution of \eqref{eq:4.6.7} is

\begin{eqnarray*}
{\bf y} = c_1 e^t \left[ \begin{array} \\ \sin 3t + 5 \cos 3t \\ 2 \cos 3t \end{array} \right] + c_2 e^t \left[ \begin{array} \\ -\cos 3t + 5 \sin 3t \\ 2 \sin 3t \end{array} \right].
\end{eqnarray*}

### Example $$\PageIndex{3}$$

Find the general solution of

\label{eq:4.6.8}
{\bf y}' = \left[ \begin{array} \\ {-5} & 5 & 4 \\ {-8} & 7 & 6 \\ 1 & 0 & 0 \end{array} \right] {\bf y}.

The characteristic polynomial of the coefficient matrix $$A$$ in \eqref{eq:4.6.8} is

\begin{eqnarray*}
\left| \begin{array} \\ -5 - \lambda & 5 & 4 \\ -8 & 7 - \lambda & 6 \\ \phantom{-} 1 & 0 & -\lambda \end{array} \right| = -(\lambda-2)(\lambda^2 + 1).
\end{eqnarray*}

Hence, the eigenvalues of $$A$$ are $$\lambda_1=2$$, $$\lambda_2=i$$, and $$\lambda_3=-i$$. The augmented matrix of $$(A-2I){\bf x=0}$$ is

\begin{eqnarray*}
\left[ \begin{array} \\ -7 & 5 & 4 & \vdots & 0 \\ -8 & 5 & 6 & \vdots & 0 \\ 1 & 0 & -2 & \vdots & 0 \end{array} \right],
\end{eqnarray*}

which is row equivalent to

\begin{eqnarray*}
\left[ \begin{array} \\ 1 & 0 & -2 & \vdots & 0 \\ 0 & 1 & -2 & \vdots & 0 \\ 0 & 0 & 0 & \vdots & 0 \end{array} \right].
\end{eqnarray*}

Therefore $$x_1=x_2=2x_3$$. Taking $$x_3=1$$ yields

\begin{eqnarray*}
{\bf x}_1 = \left[ \begin{array} \\ 2 \\ 2 \\ 1 \end{array} \right],
\end{eqnarray*}

so

\begin{eqnarray*}
{\bf y}_1 = \left[ \begin{array} \\ 2 \\ 2 \\ 1 e^{2t} \end{array} \right]
\end{eqnarray*}

is a solution of \eqref{eq:4.6.8}.

The augmented matrix of $$(A-iI){\bf x=0}$$ is

\begin{eqnarray*}
\left[ \begin{array} \\ -5-i & 5 & 4 & \vdots & 0 \\ -8 & 7-i & 6 & \vdots & 0 \\ \phantom{-} 1 & 0 & -i & \vdots & 0 \end{array} \right],
\end{eqnarray*}

which is row equivalent to

\begin{eqnarray*}
\left[ \begin{array} \\ 1 & 0 & -i & \vdots & 0 \\ 0 & 1 & 1-i & \vdots & 0 \\ 0 & 0 & 0 & \vdots & 0 \end{array} \right].
\end{eqnarray*}

Therefore $$x_1=ix_3$$ and $$x_2=-(1-i)x_3$$. Taking $$x_3=1$$ yields the eigenvector

\begin{eqnarray*}
{\bf x}_2 = \left[ \begin{array} \\ i \\ -1+i \\ 1 \end{array} \right].
\end{eqnarray*}

The real and imaginary parts of

\begin{eqnarray*}
(\cos t+i \sin t) \left[ \begin{array} \\ i \\ -1+i \\ 1 \end{array} \right]
\end{eqnarray*}

are

\begin{eqnarray*}
{\bf y}_2 = \left[ \begin{array} \\ -\sin t \\ -\cos t - \sin t \\ \cos t \end{array} \right] \quad \mbox{and} \quad {\bf y}_3 = \left[ \begin{array} \\ \cos t \\ \cos t - \sin t \\ \sin t \end{array} \right],
\end{eqnarray*}

which are solutions of \eqref{eq:4.6.8}. Since the Wronskian of $$\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$$ at $$t=0$$ is

\begin{eqnarray*}
\left| \begin{array} \\ 2 & 0 & 1 \\ 2 & -1 & 1 \\ 1 & 1 & 0 \end{array} \right| = 1,
\end{eqnarray*}

$$\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$$ is a fundamental set of solutions of \eqref{eq:4.6.8}. The general solution of \eqref{eq:4.6.8} is

\begin{eqnarray*}
{\bf y} = c_1 \left[ \begin{array} \\ 2 \\ 2 \\ 1 e^{2t} \end{array} \right] + c_2 \left[ \begin{array} \\ -\sin t \\ -\cos t -\sin t \\ \cos t \end{array} \right] + c_3 \left[ \begin{array} \\ \cos t \\ \cos t - \sin t \\ \sin t \end{array} \right].
\end{eqnarray*}

### Example$$\PageIndex{4}$$

Find the general solution of

\label{eq:4.6.9}
{\bf y}' = \left[ \begin{array} \\ 1 & {-1} & {-2} \\ 1 & 3 & 2 \\ 1 & {-1} & 2 \end{array} \right] {\bf y}.

The characteristic polynomial of the coefficient matrix $$A$$ in \eqref{eq:4.6.9} is

\begin{eqnarray*}
\left| \begin{array} \\ 1 - \lambda & -1 & -2 \\ 1 & 3-\lambda & \phantom{-} 2 \\ 1 & -1 & 2-\lambda \end{array} \right| = -(\lambda - 2) \left( (\lambda-2)^2 + 4 \right).
\end{eqnarray*}

Hence, the eigenvalues of $$A$$ are $$\lambda_1=2$$, $$\lambda_2=2+2i$$, and $$\lambda_3=2-2i$$. The augmented matrix of $$(A-2I){\bf x=0}$$ is

\begin{eqnarray*}
\left[ \begin{array} \\ -1 & -1 & -2 & \vdots & 0 \\ 1 & 1 & 2 & \vdots & 0 \\ 1 & -1 & 0 & \vdots & 0 \end{array} \right],
\end{eqnarray*}

which is row equivalent to

\begin{eqnarray*}
\left[ \begin{array} \\ 1 & 0 & 1 & \vdots & 0 \\ 0 & 1 & 1 & \vdots & 0 \\ 0 & 0 & 0 & \vdots & 0 \end{array} \right].
\end{eqnarray*}

Therefore $$x_1=x_2=-x_3$$. Taking $$x_3=1$$ yields

\begin{eqnarray*}
{\bf x}_1 = \left[ \begin{array} \\ -1 \\ -1 \\ 1 \end{array} \right],
\end{eqnarray*}

so

\begin{eqnarray*}
{\bf y}_1 = \left[ \begin{array} \\ -1 \\ -1 \\ 1 e^{2t} \end{array} \right]
\end{eqnarray*}

is a solution of \eqref{eq:4.6.9}.

The augmented matrix of $$\left(A-(2+2i)I\right){\bf x=0}$$ is

\begin{eqnarray*}
\left[ \begin{array} \\ -1-2i & -1 & -2 & \vdots & 0 \\ 1 & 1-2i & \phantom{-} 2 & \vdots & 0 \\ 1 & -1 & -2 & \vdots & 0 \end{array} \right],
\end{eqnarray*}

which is row equivalent to

\begin{eqnarray*}
\left[ \begin{array} \\ 1 & 0 & -i & \vdots & 0 \\ 0 & 1 & i & \vdots & 0 \\ 0 & 0 & 0 & \vdots & 0 \end{array} \right].
\end{eqnarray*}

Therefore $$x_1=ix_3$$ and $$x_2=-ix_3$$. Taking $$x_3=1$$ yields the eigenvector

\begin{eqnarray*}
{\bf x}_2 = \left[ \begin{array} \\ i \\ -i \\ 1 \end{array} \right]
\end{eqnarray*}

The real and imaginary parts of

\begin{eqnarray*}
e^{2t} (\cos 2t + i \sin 2t) \left[ \begin{array} \\ i \\ -i \\ 1 \end{array} \right]
\end{eqnarray*}

are

\begin{eqnarray*}
{\bf y} _2 = e^{2t} \left[ \begin{array} \\ -\sin 2t \\ \sin 2t \\ \cos 2t \end{array} \right] \quad \mbox{and} \quad {\bf y}_2 = e^{2t} \left[ \begin{array} \\ \cos 2t \\ -\cos 2t \\ \sin 2t \end{array} \right],
\end{eqnarray*}

which are solutions of \eqref{eq:4.6.9}. Since the Wronskian of $$\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$$ at $$t=0$$ is

\begin{eqnarray*}
\left| \begin{array} \\ -1 & 0 & 1 \\ -1 & 0 & -1 \\ 1 & 1 & 0 \end{array} \right| = -2,
\end{eqnarray*}

$$\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$$ is a fundamental set of solutions of \eqref{eq:4.6.9}. The general solution of \eqref{eq:4.6.9} is

\begin{eqnarray*}
{\bf y} = c_1 \left[ \begin{array} \\ -1 \\ -1 \\ 1 e^{2t} \end{array} \right] + c_2 e^{2t} \left[ \begin{array} \\ -\sin 2t \\ \sin 2t \\ \cos 2t \end{array} \right] + c_3 e^{2t} \left[ \begin{array} \\ \cos 2t \\ -\cos 2t \\ \sin 2t \end{array} \right].
\end{eqnarray*}

## Geometric Properties of Solutions when $$n=2$$

We'll now consider the geometric properties of solutions of a $$2\times2$$ constant coefficient system

\label{eq:4.6.10}
\left[ \begin{array} \\ {y_1'} \\ {y_2'} \end{array} \right] = \left[\begin{array}{cc}a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right] \left[ \begin{array} \\ {y_1} \\ {y_2} \end{array} \right]

under the assumptions of this section; that is, when the matrix

\begin{eqnarray*}
A = \left[ \begin{array} \\ a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]
\end{eqnarray*}

has a complex eigenvalue $$\lambda = \alpha + i\beta$$ $$\beta\ne0$${\rm}) 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's 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

\begin{eqnarray*}
{\bf x}_1 = (1+ik){\bf x} = (1 + ik) ({\bf u} + i{\bf v}) = ({\bf u} - k{\bf v}) + i({\bf v} + k{\bf u})
\end{eqnarray*}

is also a $$\lambda$$-eigenvector of $$A$$, since

\begin{eqnarray*}
A {\bf x}_1 = A ((1 + ik){\bf x}) = (1 + ik) A {\bf x} = (1 + ik) \lambda {\bf x} = \lambda ((1 + ik){\bf x}) = \lambda{\bf x}_1.
\end{eqnarray*}

The real and imaginary parts of $${\bf x}_1$$ are

\label{eq:4.6.11}
{\bf u}_1={\bf u}-k{\bf v} \quad \mbox{ and } \quad {\bf v}_1={\bf v}+k{\bf u},

so

\begin{eqnarray*}
({\bf u}_1, {\bf v}_1) = ({\bf u} - k{\bf v}, {\bf v} + k{\bf u}) = -\left[ ({\bf u}, {\bf v}) k^2 + (\|{\bf v}\|^2 - \|{\bf u}\|^2 k - ({\bf u},{\bf v})\right].
\end{eqnarray*}

Therefore $$({\bf u}_1,{\bf v}_1)=0$$ if

\label{eq:4.6.12}
({\bf u},{\bf v})k^2+(\|{\bf v}\|^2-\|{\bf u}\|^2)k-({\bf u},{\bf v})=0.

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 $$(4.6E.28)$$).

### Example $$\PageIndex{5}$$:

In Example $$(4.6.1)$$, we found the eigenvector

\begin{eqnarray*}
{\bf x} - \left[ \begin{array} \\ 3+4i \\ 5 \end{array} \right] = \left[ \begin{array} \\ 3 \\ 5 \end{array} \right] + i \left[ \begin{array} \\ 4 \\ 0 \end{array} \right]
\end{eqnarray*}

for the matrix of the system \eqref{eq:4.6.6}. Here $${\bf u}=\displaystyle { \left[ \begin{array} \\ 3 \\ 5 \end{array} \right] }$$ and $${\bf v} = \left[ \begin{array} \\ 4 \\ 0 \end{array} \right]$$ are not orthogonal, since $$({\bf u},{\bf v})=12$$. Since $$\|{\bf v}\|^2-\|{\bf u}\|^2=-18$$, \eqref{eq:4.6.12} is equivalent to

\begin{eqnarray*}
2k^2 -3k -1 = 0.
\end{eqnarray*}

The zeros of this equation are $$k_1=2$$ and $$k_2=-1/2$$. Letting $$k=2$$ in \eqref{eq:4.6.11} yields

\begin{eqnarray*}
{\bf u}_1 = {\bf u}-2{\bf v} = \left[ \begin{array} \\ -5 \\ \phantom{-} 5 \end{array} \right] \quad \mbox{and} \quad {\bf v}_1 = {\bf v} + 2 {\bf u} = \left[ \begin{array} \\ 10 \\ 10 \end{array} \right],
\end{eqnarray*}

and $$({\bf u}_1,{\bf v}_1)=0$$. Letting $$k=-1/2$$ in \eqref{eq:4.6.11} yields

\begin{eqnarray*}
{\bf u}_1 = {\bf u} + {{\bf v} \over 2} = \left[ \begin{array} \\ 5 \\ 5 \end{array} \right] \quad \mbox{and} \quad {\bf v}_1 = {\bf v} - {{\bf u} \over 2} = {1 \over 2} \left[ \begin{array} \\ -5 \\ \phantom{-} 5 \end{array} \right],
\end{eqnarray*}

and again $$({\bf u}_1,{\bf v}_1)=0$$.

(The numbers don't always work out as nicely as in this example. You'll need a calculator or computer to do Exercises $$(4.6E.29)$$ to $$(4.6E.40)$$.)

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 \eqref{eq:4.6.5}, the solutions of \eqref{eq:4.6.10} are given by

\label{eq:4.6.13} {\bf y}=e^{\alpha t}\left[(c_1\cos\beta
t+c_2\sin\beta t){\bf u} +(-c_1\sin\beta t+c_2\cos\beta t){\bf
v}\right].

For convenience, let's call the curve traversed by $$e^{-\alpha t}{\bf y}(t)$$ a $$\textcolor{blue}{\mbox{shadow trajectory}}$$ of \eqref{eq:4.6.10}. Multiplying \eqref{eq:4.6.13} by $$e^{-\alpha t}$$ yields

\begin{eqnarray*}
e^{-\alpha t} {\bf y} (t) = z_1(t) {\bf U} + z_2 (t) {\bf V},
\end{eqnarray*}

where

\begin{eqnarray*}
z_1(t)&=&\|{\bf u}\|(c_1\cos\beta t+c_2\sin\beta
t)\\ z_2(t)&=&\|{\bf v}\|(-c_1\sin\beta t+c_2\cos\beta t).
\end{eqnarray*}

Therefore

\begin{eqnarray*}
{(z_1(t))^2 \over \| {\bf u} \|^2} + {(z_2(t))^2 \over \| {\bf v} \|^2} = c_1^2 + c_2^2
\end{eqnarray*}

(verify!), which means that the shadow trajectories of \eqref{eq:4.6.10} are ellipses centered at the origin, with axes of symmetry parallel to $${\bf U}$$ and $${\bf V}$$. Since

\begin{eqnarray*}
z_1' = {\beta \| {\bf u} \| \over \| {\bf v} \|} z_2 \quad \mbox{and} \quad z_2' = -{\beta \| {\bf v} \| \over \| {\bf u} \|} z_1,
\end{eqnarray*}

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 $$4.6.1$$ and $$4.6.2$$).

### Figure $$4.6.2$$

If $$\alpha=0$$, then any trajectory of \eqref{eq:4.6.10} is a shadow trajectory of \eqref{eq:4.6.10}; therefore, if $$\lambda$$ is purely imaginary, then the trajectories of \eqref{eq:4.6.10} are ellipses traversed periodically as indicated in Figures $$4.6.1$$ and $$4.6.2$$.

If $$\alpha>0$$, then

\begin{eqnarray*}
\lim_{t\to\infty} \| {\bf y}(t) \| = \infty \quad \mbox{and} \quad \lim_{t\to-\infty} {\bf y}(t) = 0,
\end{eqnarray*}

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 $$4.6.3$$ and $$4.6.4$$.

If $$\alpha<0$$, then

\begin{eqnarray*}
\lim_{t\to-\infty} \| {\bf y}(t) \| = \infty \quad \mbox{and} \quad \lim_{t\to\infty}{\bf y}(t) = 0,
\end{eqnarray*}

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 $$4.6.5$$ and $$4.6.6$$.

### Figure $$4.6.3$$

$$\alpha>0$$; shadow trajectory spiraling outward

### Figure $$4.6.4$$

$$\alpha>0$$; shadow trajectory spiraling outward

### Figure $$4.6.5$$

$$\alpha<0$$; shadow trajectory spiraling inward

### Figure $$4.6.6$$

$$\alpha<0$$; shadow trajectory spiraling inward