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# 4.5: Constant Coefficient Homogeneous Systems II

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## Constant Coefficient Homogeneous Systems II

We saw in Section 4.4 that if an $$n\times n$$ constant matrix $$A$$ has $$n$$ real eigenvalues $$\lambda_1$$, $$\lambda_2$$, $$\dots$$, $$\lambda_n$$ (which need not be distinct) with associated linearly independent eigenvectors $${\bf x}_1$$, $${\bf x}_2$$, $$\dots$$, $${\bf x}_n$$, then the general solution of $${\bf y}'=A{\bf y}$$ is

\begin{eqnarray*}
{\bf y} = c_1{\bf x}_1 e^{\lambda_1 t} = c_2{\bf x}_2 e^{\lambda_2 t} + \cdots + c_n{\bf x}_n e^{\lambda_n t}.
\end{eqnarray*}

In this section we consider the case where $$A$$ has $$n$$ real eigenvalues, but does not have $$n$$ linearly independent eigenvectors. It is shown in linear algebra that this occurs if and only if $$A$$ has at least one eigenvalue of multiplicity $$r>1$$ such that the associated eigenspace has dimension less than $$r$$. In this case $$A$$ is said to be $$\textcolor{blue}{\mbox{defective}}$$. Since it's beyond the scope of this book to give a complete analysis of systems with defective coefficient matrices, we will restrict our attention to some commonly occurring special cases.

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

Show that the system

\label{eq:4.5.1}
{\bf y}'= \left[ \begin{array} \\ {11} & {-25} \\ 4 & {-9} \end{array} \right] {\bf y}

does not have a fundamental set of solutions of the form $$\{{\bf x}_1e^{\lambda_1t},{\bf x}_2e^{\lambda_2t}\}$$, where $$\lambda_1$$ and $$\lambda_2$$ are eigenvalues of the coefficient matrix $$A$$ of \eqref{eq:4.5.1} and $${\bf x}_1$$, and $${\bf x}_2$$ are associated linearly independent eigenvectors.

The characteristic polynomial of $$A$$ is

\begin{eqnarray*}
\left[ \begin{array} \\ {11} -\lambda & {-25} \\ 4 & {-9} -\lambda \end{array} \right]
&=&(\lambda-11)(\lambda+9)+100\\
&=&\lambda^2-2\lambda+1=(\lambda-1)^2.
\end{eqnarray*}

Hence, $$\lambda=1$$ is the only eigenvalue of $$A$$. The augmented matrix of the system $$(A-I){\bf x}={\bf 0}$$ is

\begin{eqnarray*}
\left[ \begin{array} \\ 10 & -25 & \vdots & 0 \\ 4 & -10 & \vdots \$ 0 \end{array} \right],
\end{eqnarray*}

which is row equivalent to

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

Hence, $$x_1=5x_2/2$$ where $$x_2$$ is arbitrary. Therefore all eigenvectors of $$A$$ are scalar multiples of $${\bf x}_1 = \left[ \begin{array} \\ 5 \\ 2 \end{array} \right]$$, so $$A$$ does not have a set of two linearly independent eigenvectors.

From Example $$(4.5.1)$$, we know that all scalar multiples of $${\bf y}_1 = \left[ \begin{array} \\ 5 \\ 2 \end{array} \right] e^t$$ are solutions of \eqref{eq:4.5.1}; however, to find the general solution we must find a second solution $${\bf y}_2$$ such that $$\{{\bf y}_1,{\bf y}_2\}$$ is linearly independent. Based on your recollection of the procedure for solving a constant coefficient scalar equation

\begin{eqnarray*}
ay'' + by' + cy = 0
\end{eqnarray*}

in the case where the characteristic polynomial has a repeated root, you might expect to obtain a second solution of \eqref{eq:4.5.1} by multiplying the first solution by $$t$$. However, this yields $${\bf y}_2 = \left[ \begin{array} \\ 5 \\ 2 \end{array} \right] te^t$$, which doesn't work, since

\begin{eqnarray*}
{\bf y}_2' = \left[ \begin{array} \\ 5 \\ 2 \end{array} \right] (t e^t + e^t), \quad \mbox{while} \quad \left[ \begin{array} \\ 11 & -25 \\ 4 & -9 \end{array} \right] {\bf y}_2 = \left[ \begin{array} \\ 5 \\ 2 \end{array} \right] t e^t.
\end{eqnarray*}

The next theorem shows what to do in this situation.

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

Suppose the $$n\times n$$ matrix $$A$$ has an eigenvalue $$\lambda_1$$ of multiplicity $$\ge2$$ and the associated eigenspace has dimension $$1;$$ that is, all $$\lambda_1$$-eigenvectors of $$A$$ are scalar multiples of an eigenvector $${\bf x}$$. Then there are infinitely many vectors $${\bf u}$$ such that

\label{eq:4.5.2}
(A-\lambda_1I){\bf u}={\bf x}.

Moreover, if $${\bf u}$$ is any such vector then

\label{eq:4.5.3}

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

Proof

Add proof here and it will automatically be hidden if you have a "AutoNum" template active on the page.

A complete proof of this theorem is beyond the scope of this book. The difficulty is in proving that there's a vector $${\bf u}$$ satisfying \eqref{eq:4.5.2}, since $$\det(A-\lambda_1I)=0$$. We'll take this without proof and verify the other assertions of the theorem.

We already know that $${\bf y}_1$$ in \eqref{eq:4.5.3} is a solution of $${\bf y}'=A{\bf y}$$. To see that $${\bf y}_2$$ is also a solution, we compute

\begin{eqnarray*}
{\bf y}_2'-A{\bf y}_2&=&\lambda_1{\bf u}e^{\lambda_1t}+{\bf x} e^{\lambda_1t}
+\lambda_1{\bf x} te^{\lambda_1t}-A{\bf u}e^{\lambda_1t}-A{\bf x} te^{\lambda_1t}\\
&=&(\lambda_1{\bf u}+{\bf x} -A{\bf u})e^{\lambda_1t}+(\lambda_1{\bf x} -A{\bf x} )te^{\lambda_1t}.
\end{eqnarray*}

Since $$A{\bf x}=\lambda_1{\bf x}$$, this can be written as

\begin{eqnarray*}
{\bf y}_2' - A{\bf y}_2 = - \left( (A -\lambda_1 l) {\bf u} - {\bf x} \right) e^{\lambda_1 t},
\end{eqnarray*}

and now \eqref{eq:4.5.2} implies that $${\bf y}_2'=A{\bf y}_2$$.

To see that $${\bf y}_1$$ and $${\bf y}_2$$ are linearly independent, suppose $$c_1$$ and $$c_2$$ are constants such that

\label{eq:4.5.4}
c_1{\bf y}_1+c_2{\bf y}_2=c_1{\bf x}e^{\lambda_1t}+c_2({\bf u}e^{\lambda_1t} +{\bf x}te^{\lambda_1t})={\bf 0}.

We must show that $$c_1=c_2=0$$. Multiplying \eqref{eq:4.5.4} by $$e^{-\lambda_1t}$$ shows that

\label{eq:4.5.5}
c_1{\bf x}+c_2({\bf u} +{\bf x}t)={\bf 0}.

By differentiating this with respect to $$t$$, we see that $$c_2{\bf x}={\bf 0}$$, which implies $$c_2=0$$, because $${\bf x}\ne{\bf 0}$$. Substituting $$c_2=0$$ into \eqref{eq:4.5.5} yields $$c_1{\bf x}={\bf 0}$$, which implies that $$c_1=0$$, again because $${\bf x}\ne{\bf 0}$$

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

Use Theorem $$(4.5.1)$$ to find the general solution of the system

\label{eq:4.5.6}
{\bf y}' = \left[ \begin{array} \\ {11} & {-25} \\ 4 & {-9} \end{array} \right] {\bf y}

considered in Example $$(4.5.1)$$.

In Example $$(4.5.1)$$ we saw that $$\lambda_1=1$$ is an eigenvalue of multiplicity $$2$$ of the coefficient matrix $$A$$ in \eqref{eq:4.5.6}, and that all of the eigenvectors of $$A$$ are multiples of

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

Therefore

\begin{eqnarray*}
{\bf y}_1 = \left[ \begin{array} \\ 5 \\ 2 \end{array} \right] e^t
\end{eqnarray*}

is a solution of \eqref{eq:4.5.6}. From Theorem $$(4.5.1)$$, a second solution is given by $${\bf y}_2={\bf u}e^t+{\bf x}te^t$$, where $$(A-I){\bf u}={\bf x}$$. The augmented matrix of this system is

\begin{eqnarray*}
\left[ \begin{array} \\ 10 & -25 & \vdots & 5 \\ 4 & -10 & \vdots & 2 \end{array} \right],
\end{eqnarray*}

which is row equivalent to

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

Therefore the components of $${\bf u}$$ must satisfy

\begin{eqnarray*}
u_1 - {5 \over 2} u_2 = {1 \over 2},
\end{eqnarray*}

where $$u_2$$ is arbitrary. We choose $$u_2=0$$, so that $$u_1=1/2$$ and

\begin{eqnarray*}
{\bf u} = \left[ \begin{array} \\ {1 \over 2} \\ 0 \end{array} \right].
\end{eqnarray*}

Thus,

\begin{eqnarray*}
{\bf y}_2 = \left[ \begin{array} \\ 1 \\ 0 \end{array} \right] {e^t \over 2} + \left[ \begin{array} \\ 5 \\ 2 \end{array} \right] e^t.
\end{eqnarray*}

Since $${\bf y}_1$$ and $${\bf y}_2$$ are linearly independent by Theorem $$(4.5.1)$$, they form a fundamental set of solutions of \eqref{eq:4.5.6}. Therefore the general solution of \eqref{eq:4.5.6} is

\begin{eqnarray*}
{\bf y} = c_1 \left[ \begin{array} \\ 5 \\ 2 \end{array} \right] e^t + c_2 \left( \left[ \begin{array} \\ 1 \\ 0 \end{array} \right] {e^t \over 2} + \left[ \begin{array} \\ 5 \\ 2 \end{array} \right] t e^t \right)
\end{eqnarray*}

Note that choosing the arbitrary constant $$u_2$$ to be nonzero is equivalent to adding a scalar multiple of $${\bf y}_1$$ to the second solution $${\bf y}_2$$ (Exercise $$(4.5E.33)$$).

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

Find the general solution of

\label{eq:4.5.7}
{\bf y}' = \left[ \begin{array} \\ 3 & 4 & {-10} \\ 2 & 1 & {-2} \\ 2 & 2 & {-5} \end{array} \right] {\bf y}.

The characteristic polynomial of the coefficient matrix $$A$$ in \eqref{eq:4.5.7} is

\begin{eqnarray*}
\left[ \begin{array} \\ 3-\lambda & 4 & -10 \\ 2 & 1-\lambda & -2 \\ 2 & 2 & -5-\lambda \end{array} \right] = -(\lambda-1)(\lambda+1)^2.
\end{eqnarray*}

Hence, the eigenvalues are $$\lambda_1=1$$ with multiplicity $$1$$ and $$\lambda_2=-1$$ with multiplicity $$2$$.

Eigenvectors associated with $$\lambda_1=1$$ must satisfy $$(A-I){\bf x}={\bf 0}$$. The augmented matrix of this system is

\begin{eqnarray*}
\left[ \begin{array} \\ 2 & 4 & -10 & \vdots & 0 \\ 2 & 0 & -2 & \vdots & 0 \\ 2 & 2 & -6 & \vdots & 0 \end{array} \right]
\end{eqnarray*}

which is row equivalent to

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

Hence, $$x_1 =x_3$$ and $$x_2 =2 x_3$$, where $$x_3$$ is arbitrary. Choosing $$x_3=1$$ yields the eigenvector

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

Therefore

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

is a solution of \eqref{eq:4.5.7}.

Eigenvectors associated with $$\lambda_2 =-1$$ satisfy $$(A+I){\bf x}={\bf 0}$$. The augmented matrix of this system is

\begin{eqnarray*}
\left[ \begin{array} \\ 4 & 4 & -10 & \vdots & 0 \\ 2 & 2 & -2 & \vdots & 0 \\ 2 & 2 & -4 & \vdots & 0 \end{array} \right],
\end{eqnarray*}

which is row equivalent to

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

Hence, $$x_3=0$$ and $$x_1 =-x_2$$, where $$x_2$$ is arbitrary. Choosing $$x_2=1$$ yields the eigenvector

\begin{eqnarray*}
{\bf y}_2 = \left[ \begin{array} \\ -1 \\ 1 \\ 0 \end{array} \right],
\end{eqnarray*}

so

\begin{eqnarray*}
{\bf y}_2 = \left[ \begin{array} \\ -1 \\ 1 \\ 0 \end{array} \right] e^{-t}
\end{eqnarray*}

is a solution of \eqref{eq:4.5.7}.

Since all the eigenvectors of $$A$$ associated with $$\lambda_2=-1$$ are multiples of $${\bf x}_2$$, we must now use Theorem $$(4.5.1)$$ to find a third solution of \eqref{eq:4.5.7} in the form

\label{eq:4.5.8}
{\bf y}_3={\bf u}e^{-t} + \left[ \begin{array} \\ {-1} \\ 1 \\ 0 \end{array} \right] te^{-t},

where $${\bf u}$$ is a solution of $$(A+I){\bf u=x}_2$$. The augmented matrix of this system is

\begin{eqnarray*}
\left[ \begin{array} \\ 4 & 4 & -10 & \vdots & -1 \\ 2 & 2 & -2 & \vdots & 1 \\ 2 & 2 & -4 & \vdots & 0 \end{array} \right],
\end{eqnarray*}

which is row equivalent to

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

Hence, $$u_3=1/2$$ and $$u_1 =1-u_2$$, where $$u_2$$ is arbitrary. Choosing $$u_2=0$$ yields

\begin{eqnarray*}
{\bf u} = \left[ \begin{array} \\ 1 \\ 0 \\ {1 \over 2} \end{array} \right],
\end{eqnarray*}

and substituting this into \eqref{eq:4.5.8} yields the solution

\begin{eqnarray*}
{\bf y}_3 = \left[ \begin{array} \\ 2 \\ 0 \\ 1 \end{array} \right] {e^{-t} \over 2} + \left[ \begin{array} \\ -1 \\ 1 \\ 0 \end{array} \right] t e^{-t}
\end{eqnarray*}

of \eqref{eq:4.5.7}.

Since the Wronskian of $$\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$$ at $$t=0$$ is

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

$$\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$$ is a fundamental set of solutions of \eqref{eq:4.5.7}. Therefore the general solution of \eqref{eq:4.5.7} is

\begin{eqnarray*}
{\bf y} = c_1 \left[ \begin{array} \\ 1 \\ 2 \\ 1 \end{array} \right] e^t + c_2 \left[ \begin{array} \\ -1 \\ 1 \\ 0 \end{array} \right] e^{-t} + c_3 \left( \left[ \begin{array} \\ 2 \\ 0 \\ 1 \end{array} \right] {e^{-t} \over 2} + \left[ \begin{array} \\ -1 \\ 1 \\ 0 \end{array} \right] e^{-t} \right).
\end{eqnarray*}

### Theorem $$\PageIndex{2}$$

Suppose the $$n\times n$$ matrix $$A$$ has an eigenvalue $$\lambda_1$$ of multiplicity $$\ge 3$$ and the associated eigenspace is one-dimensional; that is, all eigenvectors associated with $$\lambda_1$$ are scalar multiples of the eigenvector $${\bf x}.$$ Then there are infinitely many vectors $${\bf u}$$ such that

\label{eq:4.5.9}
(A-\lambda_1I){\bf u}={\bf x},

and, if $${\bf u}$$ is any such vector, there are infinitely many vectors $${\bf v}$$ such that

\label{eq:4.5.10}
(A-\lambda_1I){\bf v}={\bf u}.

If $${\bf u}$$ satisfies {\rm\eqref{eq:4.5.9}} and $${\bf v}$$ satisfies \eqref{eq:4.5.10}, then

\begin{eqnarray*}
{\bf y}_1 &=&{\bf x} e^{\lambda_1t},\\
{\bf y}_2&=&{\bf u}e^{\lambda_1t}+{\bf x} te^{\lambda_1t},\mbox{
and }\\
{\bf y}_3&=&{\bf v}e^{\lambda_1t}+{\bf u}te^{\lambda_1t}+{\bf
x} {t^2e^{\lambda_1t}\over2}
\end{eqnarray*}

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

Proof

Add proof here and it will automatically be hidden if you have a "AutoNum" template active on the page.

Again, it's beyond the scope of this book to prove that there are vectors $${\bf u}$$ and $${\bf v}$$ that satisfy \eqref{eq:4.5.9} and
\eqref{eq:4.5.10}. Theorem $$(4.5.1)$$ implies that $${\bf y}_1$$ and $${\bf y}_2$$ are solutions of $${\bf y}'=A{\bf y}$$. We leave the rest of the proof to you (Exercise $$(4.5E.34)$$).

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

Use Theorem $$(4.5.2)$$ to find the general solution of

\label{eq:4.5.11}
{\bf y}' = \left[ \begin{array} \\ 1 & 1 & 1 \\ 1 & 3 & {-1} \\ 0 & 2 & 2 \end{array} \right] {\bf y}.

The characteristic polynomial of the coefficient matrix $$A$$ in \eqref{eq:4.5.11} is

\begin{eqnarray*}
\left| \begin{array} \\ 1-\lambda & 1 & \phantom{-}1 \\ 1 & 3-\lambda & -1 \\ 0 & 2 & -\lambda \end{array} \right| = -(\lambda-2)^3.
\end{eqnarray*}

Hence, $$\lambda_1=2$$ is an eigenvalue of multiplicity $$3$$. The associated eigenvectors satisfy $$(A-2I){\bf x=0}$$. The augmented matrix of this system is

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

Hence, $$x_1 =x_3$$ and $$x_2 = 0$$, so the eigenvectors are all scalar multiples of

\begin{eqnarray*}
{\bf x}_1 = \left[ \begin{array} \\ 1 \\ 0 \\ 1 \end{array} \right].
\end{eqnarray*}

Therefore

\begin{eqnarray*}
{\bf y}_1 = \left[ \begin{array} \\ 1 \\ 0 \\ 1 \end{array} \right] e^{2t}
\end{eqnarray*}

is a solution of \eqref{eq:4.5.11}.

We now find a second solution of \eqref{eq:4.5.11} in the form

\begin{eqnarray*}
{\bf y}_2 = {\bf u} e^{2t} + \left[ \begin{array} \\ 1 \\ 0 \\ 1 \end{array} \right] t e^{2t},
\end{eqnarray*}

where $${\bf u}$$ satisfies $$(A-2I){\bf u=x}_1$$. The augmented matrix of this system is

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

which is row equivalent to

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

Letting $$u_3=0$$ yields $$u_1=-1/2$$ and $$u_2=1/2$$; hence,

\begin{eqnarray*}
{\bf u} = {1 \over 2} \left[ \begin{array} \\ -1 \\ 1 \\ 0 \end{array} \right]
\end{eqnarray*}

and

\begin{eqnarray*}
{\bf y}_2 = \left[ \begin{array} \\ -1 \\ 1 \\ 0 \end{array} \right] {e^{2t} \over 2} + \left[ \begin{array} \\ 1 \\ 0 \\ 1 \end{array} \right] t e^{2t}
\end{eqnarray*}

is a solution of \eqref{eq:4.5.11}.

We now find a third solution of \eqref{eq:4.5.11} in the form

\begin{eqnarray*}
{\bf y}_3 = {\bf v} e^{2t} + \left[ \begin{array} \\ -1 \\ 1 \\ 0 \end{array} \right] {t e^{2t} \over 2} + \left[ \begin{array} \\ 1 \\ 0 \\ 1 \end{array} \right] {t^2 e^{2t} \over 2}
\end{eqnarray*}

where $${\bf v}$$ satisfies $$(A-2I){\bf v}={\bf u}$$. The augmented matrix of this system is

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

which is row equivalent to

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

Letting $$v_3=0$$ yields $$v_1=1/2$$ and $$v_2=0$$; hence,

\begin{eqnarray*}
{\bf v} = {1 \over 2} \left[ \begin{array} \\ 1 \\ 0 \\ 0 \end{array} \right].
\end{eqnarray*}

Therefore

\begin{eqnarray*}
{\bf y}_3 = \left[ \begin{array} \\ 1 \\ 0 \\ 0 \end{array} \right] {e^{2t} \over 2} + \left[ \begin{array} \\ -1 \\ 1 \\ 0 \end{array} \right] {t e^{2t} \over 2} + \left[ \begin{array} \\ 1 \\ 0 \\ 1 \end{array} \right] {t^2 e^{2t} \over 2}
\end{eqnarray*}

is a solution of \eqref{eq:4.5.11}. Since $${\bf y}_1$$, $${\bf y}_2$$, and $${\bf y}_3$$ are linearly independent by Theorem $$(4.5.2)$$, they form a fundamental set of solutions of \eqref{eq:4.5.11}. Therefore the general solution of \eqref{eq:4.5.11} is

\begin{eqnarray*}
{\bf y} &=& c_1 \left[ \begin{array} \\ 1 \\ 0 \\ 1 \end{array} \right] e^{2t} + c_2 \left( \left[ \begin{array} \\ {-1} \\ 1 \\ 0 \end{array} \right] {e^{2t}\over2} + \left[ \begin{array} \\ 1 \\ 0 \\ 1 \end{array} \right] te^{2t} \right)\\
&& + c_3 \left( \left[ \begin{array} \\ 1 \\ 0 \\ 0 \end{array} \right] {e^{2t}\over2} + \left[ \begin{array} \\ {-1} \\ 1 \\ 0 \end{array} \right] {te^{2t}\over2} + \left[ \begin{array} \\ 1 \\ 0 \\ 1 \end{array} \right] {t^2e^{2t}\over2} \right).
\end{eqnarray*}

### Theorem $$\PageIndex{3}$$

Suppose the $$n\times n$$ matrix $$A$$ has an eigenvalue $$\lambda_1$$ of multiplicity $$\ge 3$$ and the associated eigenspace is two-dimensional; that is, all eigenvectors of $$A$$ associated with $$\lambda_1$$ are linear combinations of two linearly independent eigenvectors $${\bf x}_1$$ and $${\bf x}_2$$. Then there are constants $$\alpha$$ and $$\beta$$ (not both zero) such that if

\label{eq:4.5.12}
{\bf x}_3=\alpha{\bf x}_1+\beta{\bf x}_2,

then there are infinitely many vectors $${\bf u}$$ such that

\label{eq:4.5.13}
(A-\lambda_1I){\bf u}={\bf x}_3.

If $${\bf u}$$ satisfies \eqref{eq:4.5.13}, then

\begin{eqnarray}
{\bf y}_1&=&{\bf x}_1 e^{\lambda_1t},\nonumber\\
{\bf y}_2&=&{\bf x}_2e^{\lambda_1t},\mbox{and }\nonumber\\
{\bf y}_3&=&{\bf u}e^{\lambda_1t}+{\bf x}_3te^{\lambda_1t}\label{eq:4.5.14},
\end{eqnarray}

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

Proof

We omit the proof of this theorem.

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

Use Theorem $$(4.5.3)$$ to find the general solution of

\label{eq:4.5.15}
{\bf y}' = \left[ \begin{array} \\ 0 & 0 & 1 \\ {-1} & 1 & 1 \\ {-1} & 0 & 2 \end{array} \right] {\bf y}.

The characteristic polynomial of the coefficient matrix $$A$$ in \eqref{eq:4.5.15} is

\begin{eqnarray*}
\left| \begin{array} \\ -\lambda & 0 & 1 \\ -1 & 1-\lambda & 1 \\ -1 & 0 & 2-\lambda \end{array} \right| = -(\lambda-1)^3.
\end{eqnarray*}

Hence, $$\lambda_1=1$$ is an eigenvalue of multiplicity $$3$$. The associated eigenvectors satisfy $$(A-I){\bf x=0}$$. The augmented matrix of this system is

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

which is row equivalent to

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

Hence, $$x_1 =x_3$$ and $$x_2$$ is arbitrary, so the eigenvectors are of the form

\begin{eqnarray*}
{\bf x}_1 = \left[ \begin{array} \\ x_3 \\ x_2 \\ x_3 \end{array} \right] = x_3 \left[ \begin{array} \\ 1 \\ 0 \\ 1 \end{array} \right] + x_2 \left[ \begin{array} \\ 0 \\ 1 \\ 0 \end{array} \right].
\end{eqnarray*}

Therefore the vectors

\label{eq:4.5.16}
{\bf x}_1 = \left[ \begin{array} \\ 1 \\ 0 \\ 1 \end{array} \right] \quad \mbox{and} \quad {\bf x}_2 = \left[ \begin{array} \\ 0 \\ 1 \\ 0 \end{array} \right]

form a basis for the eigenspace, and

\begin{eqnarray*}
{\bf y}_1 = \left[ \begin{array} \\ 1 \\ 0 \\ 1 \end{array} \right] e^t \quad \mbox{and} \quad {\bf y}_2 = \left[ \begin{array} \\ 0 \\ 1 \\ 0 \end{array} \right] e^t
\end{eqnarray*}

are linearly independent solutions of \eqref{eq:4.5.15}.

To find a third linearly independent solution of \eqref{eq:4.5.15}, we must find constants $$\alpha$$ and $$\beta$$ (not both zero) such that the system

\label{eq:4.5.17}
(A-I){\bf u}=\alpha{\bf x}_1+\beta{\bf x}_2

has a solution $${\bf u}$$. The augmented matrix of this system is

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

which is row equivalent to

\label{eq:4.5.18}
\left[\begin{array}{rrrcr} 1 & 0 &- 1 &\vdots& -\alpha\\ 0 & 0 & 0
&\vdots&\beta-\alpha\\ 0 & 0 & 0 &\vdots&0\end{array}
\right].

Therefore \eqref{eq:4.5.17} has a solution if and only if $$\beta=\alpha$$, where $$\alpha$$ is arbitrary. If $$\alpha=\beta=1$$ then \eqref{eq:4.5.12} and \eqref{eq:4.5.16} yield

\begin{eqnarray*}
{\bf x}_3 = {\bf x}_1 + {\bf x}_2 = \left[ \begin{array} \\ 1 \\ 0 \\ 1 \end{array} \right] + \left[ \begin{array} \\ 0 \\ 1 \\ 0 \end{array} \right] = \left[ \begin{array} \\ 1 \\ 1 \\ 1 \end{array} \right],
\end{eqnarray*}

and the augmented matrix \eqref{eq:4.5.18} becomes

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

This implies that $$u_1=-1+u_3$$, while $$u_2$$ and $$u_3$$ are arbitrary. Choosing $$u_2=u_3=0$$ yields

\begin{eqnarray*}
{\bf u} = \left[ \begin{array} \\ -1 \\ 0 \\ 0 \end{array} \right].
\end{eqnarray*}

Therefore \eqref{eq:4.5.14} implies that

\begin{eqnarray*}
{\bf y}_3 = {\bf u} e^t + {\bf x}_3 t e^t = \left[ \begin{array} \\ -1 \\ 0 \\ 0 \end{array} \right] e^t + \left[ \begin{array} \\ 1 \\ 1 \\ 1 \end{array} \right] t e^t
\end{eqnarray*}

is a solution of \eqref{eq:4.5.15}. Since $${\bf y}_1$$, $${\bf y}_2$$, and $${\bf y}_3$$ are linearly independent by Theorem $$(4.5.3)$$, they form a fundamental set of solutions for \eqref{eq:4.5.15}. Therefore the general solution of \eqref{eq:4.5.15} is

\begin{eqnarray*}
{\bf y} = c_1 \left[ \begin{array} \\ 1 \\ 0 \\ 1 \end{array} \right] e^t + c_2 \left[ \begin{array} \\ 0 \\ 1 \\ 0 \end{array} \right] e^t + c_3 \left( \left[ \begin{array} \\ -1 \\ 0 \\ 0 \end{array} \right] e^t + \left[ \begin{array} \\ 1 \\ 1 \\ 1 \end{array} \right] e^t \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:4.5.19}
\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 repeated eigenvalue $$\lambda_1$$ and the associated eigenspace is one-dimensional. In this case we know from Theorem $$(4.5.1)$$ that the general solution of \eqref{eq:4.5.19} is

\label{eq:4.5.20}
{\bf y}=c_1{\bf x}e^{\lambda_1t}+c_2({\bf u}e^{\lambda_1t}+{\bf
x}te^{\lambda_1t}),

where $${\bf x}$$ is an eigenvector of $$A$$ and $${\bf u}$$ is any one of the infinitely many solutions of

\label{eq:4.5.21}
(A-\lambda_1I){\bf u}={\bf x}.

We assume that $$\lambda_1\ne0$$.

### Figure $$4.5.1$$

Positive and negative half-planes

Let $$L$$ denote the line through the origin parallel to $${\bf x}$$. By a $$\textcolor{blue}{\mbox{half-line}}$$ of $$L$$ we mean either of the rays obtained by removing the origin from $$L$$. Equation \eqref{eq:4.5.20} is a parametric equation of the half-line of $$L$$ in the direction of $${\bf x}$$ if $$c_1>0$$, or of the half-line of |(L\) in the direction of $$-{\bf x}$$ if $$c_1<0$$. The origin is the trajectory of the trivial solution $${\bf y}\equiv{\bf 0}$$.

Henceforth, we assume that $$c_2\ne0$$. In this case, the trajectory of \eqref{eq:4.5.20} can't intersect $$L$$, since every point of $$L$$ is on a trajectory obtained by setting $$c_2=0$$. Therefore the trajectory of \eqref{eq:4.5.20} must lie entirely in one of the open half-planes bounded by $$L$$, but does not contain any point on $$L$$. Since the initial point $$(y_1(0),y_2(0))$$ defined by $${\bf y}(0)=c_1{\bf x}_1+c_2{\bf u}$$ is on the trajectory, we can determine which half-plane contains the trajectory from the sign of $$c_2$$, as shown in Figure $$4.5.1$$. For convenience we'll call the half-plane where $$c_2>0$$ the $$\textcolor{blue}{\mbox{positive half-plane}}$$. Similarly, the-half plane where $$c_2<0$$ is the $$\textcolor{blue}{\mbox{negative half-plane}}$$. You should convince yourself (Exercise $$(4.5E.35)$$) that even though there are infinitely many vectors $${\bf u}$$ that satisfy \eqref{eq:4.5.21}, they all define the same positive and negative half-planes. In the figures simply regard $${\bf u}$$ as an arrow pointing to the positive half-plane, since wen't attempted to give $${\bf u}$$ its proper length or direction in comparison with $${\bf x}$$. For our purposes here, only the relative orientation of $${\bf x}$$ and $${\bf u}$$ is important; that is, whether the positive half-plane is to the right of an observer facing the direction of $${\bf x}$$ (as in Figures $$4.5.2$$ and $$4.5.5$$), or to the left of the observer (as in Figures $$4.5.3$$ and $$4.5.4$$).

Multiplying \eqref{eq:4.5.20} by $$e^{-\lambda_1t}$$ yields

\begin{eqnarray*}
e^{-\lambda_1 t} {\bf y}(t) = c_1{\bf x} + c_2{\bf u} + c_2 t {\bf x}.
\end{eqnarray*}

Since the last term on the right is dominant when $$|t|$$ is large, this provides the following information on the direction of $${\bf y}(t)$$:

(a) Along trajectories in the positive half-plane ($$c_2>0$$), the direction of $${\bf y}(t)$$ approaches the direction of $${\bf x}$$ as $$t\to\infty$$ and the direction of $$-{\bf x}$$ as $$t\to-\infty$$.

(b) Along trajectories in the negative half-plane ($$c_2<0$$), the direction of $${\bf y}(t)$$ approaches the direction of $$-{\bf x}$$ as $$t\to\infty$$ and the direction of $${\bf x}$$ as $$t\to-\infty$$.

Since

\begin{eqnarray*}
\end{eqnarray*}

or

\begin{eqnarray*}
\end{eqnarray*}

there are four possible patterns for the trajectories of \eqref{eq:4.5.19}, depending upon the signs of $$c_2$$ and $$\lambda_1$$.
Figures $$4.5.2$$ to $$4.5.5$$ illustrate these patterns, and reveal the following principle:

If $$\lambda_1$$ and $$c_2$$ have the same sign then the direction of the trajectory approaches the direction of $$-{\bf x}$$ as $$\|{\bf y} \|\to0$$ and the direction of $${\bf x}$$ as $$\|{\bf y}\|\to\infty$$ If $$\lambda_1$$ and $$c_2$$ have opposite signs then the direction of the trajectory approaches the direction of $${\bf x}$$ as $$\|{\bf y} \|\to0$$ and the direction of $$-{\bf x}$$ as $$\|{\bf y}\|\to\infty$$.

### Figure $$4.5.2$$

Positive eigenvalue; motion away from the origin

### Figure $$4.5.3$$

Positive eigenvalue; motion away from the origin

### Figure $$4.5.4$$

Negative eigenvalue; motion toward the origin

### Figure $$4.5.5$$

Negative eigenvalue; motion toward the origin