2.6: Examples of the Matrix Method

Here we will give some examples for typical systems for the three cases mentioned in the last section.

Example 2.14. \(A=\left(\begin{array}{ll}
4 & 2 \\
3 & 3
\end{array}\right)\)

\[0=\left|\begin{array}{cc}
4-\lambda & 2 \\
3 & 3-\lambda
\end{array}\right| \label{2.44} \]

Therefore,

\[\begin{aligned}
&0=(4-\lambda)(3-\lambda)-6 \\
&0=\lambda^{2}-7 \lambda+6 \\
&0=(\lambda-1)(\lambda-6)
\end{aligned} \label{2.45} \]

The eigenvalues are then \(\lambda=1,6\). This is an example of Case I.

Eigenvectors: Next we determine the eigenvectors associated with each of these eigenvalues. We have to solve the system \(A \mathbf{v}=\lambda \mathbf{v}\) in each case.

Case \(\lambda=1\)

\[\begin{aligned}
&\left(\begin{array}{ll}
4 & 2 \\
3 & 3
\end{array}\right)\left(\begin{array}{l}
v_{1} \\
v_{2}
\end{array}\right)=\left(\begin{array}{l}
v_{1} \\
v_{2}
\end{array}\right)
\end{aligned} \label{2.46} \]

\[\begin{aligned}
&\left(\begin{array}{ll}
3 & 2 \\
3 & 2
\end{array}\right)\left(\begin{array}{l}
v_{1} \\
v_{2}
\end{array}\right)=\left(\begin{array}{l}
0 \\
0
\end{array}\right)
\end{aligned} \label{2.47} \]

This gives \(3 v_{1}+2 v_{2}=0\). One possible solution yields an eigenvector of

\(\left(\begin{array}{l}
v_{1} \\
v_{2}
\end{array}\right)=\left(\begin{array}{c}
2 \\
-3
\end{array}\right)\)

Case \(\lambda=6\).

\[\begin{aligned}
&\left(\begin{array}{ll}
4 & 2 \\
3 & 3
\end{array}\right)\left(\begin{array}{l}
v_{1} \\
v_{2}
\end{array}\right)=6\left(\begin{array}{l}
v_{1} \\
v_{2}
\end{array}\right) \\
\end{aligned} \label{2.48} \]

\[\begin{aligned}
&\left(\begin{array}{cc}
-2 & 2 \\
3 & -3
\end{array}\right)\left(\begin{array}{l}
v_{1} \\
v_{2}
\end{array}\right)=\left(\begin{array}{l}
0 \\
0
\end{array}\right)
\end{aligned} \label{2.49} \]

For this case we need to solve \(-2 v_{1}+2 v_{2}=0\). This yields

\(\left(\begin{array}{l}
v_{1} \\
v_{2}
\end{array}\right)=\left(\begin{array}{l}
1 \\
1
\end{array}\right)\)

General Solution: We can now construct the general solution.

\[\begin{aligned}
\mathbf{x}(t) &=c_{1} e^{\lambda_{1} t} \mathbf{v}_{1}+c_{2} e^{\lambda_{2} t} \mathbf{v}_{2} \\
&=c_{1} e^{t}\left(\begin{array}{c}
2 \\
-3
\end{array}\right)+c_{2} e^{6 t}\left(\begin{array}{l}
1 \\
1
\end{array}\right) \\
&=\left(\begin{array}{c}
2 c_{1} e^{t}+c_{2} e^{6 t} \\
-3 c_{1} e^{t}+c_{2} e^{6 t}
\end{array}\right)
\end{aligned} \label{2.50} \]

Example 2.15. \(A=\left(\begin{array}{ll}
3 & -5 \\
1 & -1
\end{array}\right)\).

Eigenvalues: Again, one solves the eigenvalue equation.

\[0=\left|\begin{array}{cc}
3-\lambda & -5 \\
1 & -1-\lambda
\end{array}\right| \label{2.51} \]

Therefore,

\[\begin{aligned}
&0=(3-\lambda)(-1-\lambda)+5 \\
&0=\lambda^{2}-2 \lambda+2 \\
&\lambda=\dfrac{-(-2) \pm \sqrt{4-4(1)(2)}}{2}=1 \pm i
\end{aligned} \label{2.52} \]

The eigenvalues are then \(\lambda=1+i, 1-i\). This is an example of Case III.

Eigenvectors: In order to find the general solution, we need only find the eigenvector associated with \(1+i\).

\[\begin{aligned}
\left(\begin{array}{c}
3-5 \\
1-1
\end{array}\right)\left(\begin{array}{l}
v_{1} \\
v_{2}
\end{array}\right) &=(1+i)\left(\begin{array}{l}
v_{1} \\
v_{2}
\end{array}\right) \\
\left(\begin{array}{cc}
2-i & -5 \\
1 & -2-i
\end{array}\right)\left(\begin{array}{l}
v_{1} \\
v_{2}
\end{array}\right)&=\left(\begin{array}{l}
0 \\
0
\end{array}\right)
\end{aligned} \label{2.53} \]

We need to solve \((2-i) v_{1}-5 v_{2}=0\). Thus,

\[\left(\begin{array}{l}
v_{1} \\
v_{2}
\end{array}\right)=\left(\begin{array}{c}
2+i \\
1
\end{array}\right) \label{2.54} \]

Complex Solution: In order to get the two real linearly independent solutions, we need to compute the real and imaginary parts of \(v e^{\lambda t}\).

\begin{aligned}
e^{\lambda t}\left(\begin{array}{c}
2+i \\
1
\end{array}\right) &=e^{(1+i) t}\left(\begin{array}{c}
2+i \\
1
\end{array}\right) \\
&=e^{t}(\cos t+i \sin t)\left(\begin{array}{c}
2+i \\
1
\end{array}\right) \\
&=e^{t}\left(\begin{array}{c}
(2+i)(\cos t+i \sin t) \\
\cos t+i \sin t
\end{array}\right) \\
&=e^{t}\left(\begin{array}{c}
(2 \cos t-\sin t)+i(\cos t+2 \sin t) \\
\cos t+i \sin t
\end{array}\right)+i e^{t}\left(\begin{array}{c}
\cos t+2 \sin t \\
\sin t
\end{array}\right) . \\
&=e^{t}\left(\begin{array}{c}
2 \cos t-\sin t \\
\cos t
\end{array}\right)
\end{aligned}

General Solution: Now we can construct the general solution.

\[\begin{aligned}
\mathbf{x}(t) &=c_{1} e^{t}\left(\begin{array}{l}
2 \cos t-\sin t \\
\cos t
\end{array}\right)+c_{2} e^{t}\left(\begin{array}{c}
\cos t+2 \sin t \\
\sin t
\end{array}\right) \\
&=e^{t}\left(\begin{array}{c}
c_{1}(2 \cos t-\sin t)+c_{2}(\cos t+2 \sin t) \\
c_{1} \cos t+c_{2} \sin t
\end{array}\right)
\end{aligned} \label{2.55} \]

Note: This can be rewritten as

\(\mathbf{x}(t)=e^{t} \cos t\left(\begin{array}{c}
2 c_{1}+c_{2} \\
c_{1}
\end{array}\right)+e^{t} \sin t\left(\begin{array}{c}
2 c_{2}-c_{1} \\
c_{2}
\end{array}\right)\)

Example 2.16. \(A=\left(\begin{array}{ll}
7 & -1 \\
9 & 1
\end{array}\right)\).

Eigenvalues:

\[0=\left|\begin{array}{cc}
7-\lambda & -1 \\
9 & 1-\lambda
\end{array}\right| \label{2.56} \]

Therefore,

\[\begin{aligned}
&0=(7-\lambda)(1-\lambda)+9 \\
&0=\lambda^{2}-8 \lambda+16 \\
&0=(\lambda-4)^{2}
\end{aligned} \label{2.57} \]

There is only one real eigenvalue, \(\lambda=4\). This is an example of Case II.

Eigenvectors: In this case we first solve for \(\mathbf{v}_{1}\) and then get the second linearly independent vector.

\[\begin{aligned}
&\left(\begin{array}{cc}
7 & -1 \\
9 & 1
\end{array}\right)\left(\begin{array}{l}
v_{1} \\
v_{2}
\end{array}\right)=4\left(\begin{array}{l}
v_{1} \\
v_{2}
\end{array}\right) \\
&\left(\begin{array}{ll}
3 & -1 \\
9 & -3
\end{array}\right)\left(\begin{array}{l}
v_{1} \\
v_{2}
\end{array}\right)=\left(\begin{array}{l}
0 \\
0
\end{array}\right)
\end{aligned} \label{2.58} \]

Therefore, we have

\(3 v_{1}-v_{2}=0, \quad \Rightarrow \quad\left(\begin{array}{l}
v_{1} \\
v_{2}
\end{array}\right)=\left(\begin{array}{l}
1 \\
3
\end{array}\right)\)

Second Linearly Independent Solution:

Now we need to solve \(A \mathbf{v}_{2}-\lambda \mathbf{v}_{2}=\mathbf{v}_{1}\).

\[\begin{aligned}
\left(\begin{array}{cc}7 & -1 \\ 9 & 1\end{array}\right)\left(\begin{array}{l}u_{1} \\ u_{2}\end{array}\right)-4\left(\begin{array}{l}u_{1} \\ u_{2}\end{array}\right)&=\left(\begin{array}{l}1 \\ 3\end{array}\right) \\
\left(\begin{array}{ll}3 & -1 \\ 9 & -3\end{array}\right)\left(\begin{array}{l}u_{1} \\ u_{2}\end{array}\right)&=\left(\begin{array}{l}1 \\ 3\end{array}\right)
\end{aligned} \label{2.59} \]

Expanding the matrix product, we obtain the system of equations

\[\begin{array}{r}
3 u_{1}-u_{2}=1 \\
9 u_{1}-3 u_{2}=3
\end{array} \label{2.60} \]

The solution of this system is \(\left(\begin{array}{l}u_{1} \\ u_{2}\end{array}\right)=\left(\begin{array}{l}1 \\ 2\end{array}\right)\).

General Solution: We construct the general solution as

\[\begin{aligned}
\mathbf{y}(t) &=c_{1} e^{\lambda t} \mathbf{v}_{1}+c_{2} e^{\lambda t}\left(\mathbf{v}_{2}+t \mathbf{v}_{1}\right) \\
&=c_{1} e^{4 t}\left(\begin{array}{l}
1 \\
3
\end{array}\right)+c_{2} e^{4 t}\left[\left(\begin{array}{l}
1 \\
2
\end{array}\right)+t\left(\begin{array}{l}
1 \\
3
\end{array}\right)\right] \\
&=e^{4 t}\left(\begin{array}{c}
c_{1}+c_{2}(1+t) \\
3 c_{1}+c_{2}(2+3 t)
\end{array}\right)
\end{aligned} \label{2.61} \]

2.6.1 Planar Systems - Summary

The reader should have noted by now that there is a connection between the behavior of the solutions obtained in Section 2.2 and the eigenvalues found from the coefficient matrices in the previous examples. Here we summarize some of these cases.

Type	Figure	Eigenvalues	Stability
Node		Real \(\lambda\), same signs	\(\lambda > 0\), stable
Saddle		Real \(\lambda\) opposite signs	Mostly Unstable
Center		\(\lambda\) pure imaginary	_____
Focus/Spiral Degenerate Node Line of Equilibria		Complex \(\lambda\), Re(\(\lambda \neq 0\) Repeated roots, \(\lambda > 0\), stable One zero eigenvalue	Re(\(\lambda > 0\)), stable \(\lambda > 0\), stable

Table 2.1. List of typical behaviors in planar systems.

The connection, as we have seen, is that the characteristic equation for the associated second order differential equation is the same as the eigenvalue equation of the coefficient matrix for the linear system. However, one should be a little careful in cases in which the coefficient matrix in not diagonalizable. In Table 2.2 are three examples of systems with repeated roots. The reader should look at these systems and look at the commonalities and differences in these systems and their solutions. In these cases one has unstable nodes, though they are degenerate in that there is only one accessible eigenvector.