Consider a \(2 \times 2\) system \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}\). If
we assume that \(r_{1} \neq r_{2}\), the general solution is \(\mathbf{x}=c_{1}
\xi^{(1)} e^{t_{1}^{\prime}}+c_{2} \xi^{(2)} e^{\prime 2},\) provided that
\(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly independent In this problem we
establish the linear independence of \(\xi^{(1)}\) and \(\xi^{(2)}\) by assuming
that they are linearly dependent, and then showing that this leads to a
contradiction.
$$
\begin{array}{l}{\text { (a) Note that } \xi \text { (i) satisfies the matrix
equation }\left(\mathbf{A}-r_{1} \mathbf{I}\right) \xi^{(1)}=\mathbf{0} ;
\text { similarly, note that }} \\ {\left(\mathbf{A}-r_{2} \mathbf{I}\right)
\xi^{(2)}=\mathbf{0}} \\ {\text { (b) Show that }\left(\mathbf{A}-r_{2}
\mathbf{I}\right) \xi^{(1)}=\left(r_{1}-r_{2}\right) \mathbf{\xi}^{(1)}} \\\
{\text { (c) Suppose that } \xi^{(1)} \text { and } \xi^{(2)} \text { are
linearly dependent. Then } c_{1} \xi^{(1)}+c_{2} \xi^{(2)}=\mathbf{0} \text {
and at least }}\end{array}
$$
$$
\begin{array}{l}{\text { one of } c_{1} \text { and } c_{2} \text { is not
zero; suppose that } c_{1} \neq 0 . \text { Show that }\left(\mathbf{A}-r_{2}
\mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2}
\boldsymbol{\xi}^{(2)}\right)=\mathbf{0}} \\ {\text { and also show that
}\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1}
\boldsymbol{\xi}^{(1)}+c_{2}
\boldsymbol{\xi}^{(2)}\right)=c_{1}\left(r_{1}-r_{2}\right)
\boldsymbol{\xi}^{(1)} \text { . Hence } c_{1}=0, \text { which is }} \\\
{\text { a contradiction. Therefore } \xi^{(1)} \text { and }
\boldsymbol{\xi}^{(2)} \text { are linearly independent. }}\end{array}
$$
$$
\begin{array}{l}{\text { (d) Modify the argument of part (c) in case } c_{1}
\text { is zero but } c_{2} \text { is not. }} \\ {\text { (e) Carry out a
similar argument for the case in which the order } n \text { is equal to } 3
\text { ; note that }} \\ {\text { the procedure can be extended to cover an
arbitrary value of } n .}\end{array}
$$
