Having detoured briefly to some second order examples, let me return to first order constant coefficient linear system. I want to give a new understanding to the solutions of systems of two equations that I did before in the examples. The general linear system with two equations has the following form, to remind you.
\begin{align*}
x^\prime \amp = ax + by \\
y^\prime \amp = cx + dy
\end{align*}
I can use a 2 by 2 matrix with coefficients \(a,b,c,d\) to analyze the behaviour. Then I can make the functions \(x\) and \(y\) the coordinates of a vector and write the system in terms of this matrix.
\begin{equation*}
\begin{pmatrix}
x_1^\prime \\
x_2^\prime
\end{pmatrix} = \begin{pmatrix}
a \amp b \\
c \amp d
\end{pmatrix} \begin{pmatrix}
x_1 \\
x_2
\end{pmatrix}
\end{equation*}
Let \(v\) be an eigenvector of the matrix and let \(\lambda\) be the corresponding eigenvectors.
\begin{equation*}
\begin{pmatrix}
x_1 \\
x_2
\end{pmatrix} = e^{\lambda t} v = \begin{pmatrix}
e^{\lambda t} v_1 \\
e^{\lambda t} v_1
\end{pmatrix}
\end{equation*}
To see why this is a solution, see what happens when the matrix acts on the vector.
\begin{align*}
\begin{pmatrix}
a \amp b \\
c \amp d
\end{pmatrix} \begin{pmatrix}
e^{\lambda t} v_1 \\
e^{\lambda t} v_2
\end{pmatrix} \amp = e^{\lambda t} \begin{pmatrix}
a \amp b
c \amp d
\end{pmatrix} \begin{pmatrix}
v_1
v_2
\end{pmatrix} \\
\amp = e^{\lambda t} \lambda \begin{pmatrix}
v_1
v_2
\end{pmatrix} = \begin{pmatrix}
\lambda e^{\lambda t} v_1
\lambda e^{\lambda t} v_2
\end{pmatrix} \\
\amp = \begin{pmatrix}
\dfrac{d}{dt}e^{\lambda t} v_1
\dfrac{d}{dt} e^{\lambda t} v_2
\end{pmatrix} = \frac{d}{dt} \begin{pmatrix}
e^{\lambda t} v_1
e^{\lambda t} v_2 ]
\end{pmatrix}
\end{align*}
If I allow for complex eigenvalues, the matrix has two eigenvalues, wth two linearly independent eigenvectors, giving two linearly independent solutions of this type. If the eigenvalues are real, there are exponential growth/decay solutions. If the eigenvalues are complex, these are sinusoidal solutions.
