Question

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. $$ \begin{array}{l}{\text { (a) Note that } \xi \text { (i) satisfies the matrix equation }\left(\mathbf{A}-r_{1} \mathbf{I}\right) \xi^{(1)}=\mathbf{0} ; \text { similarly, note that }} \\ {\left(\mathbf{A}-r_{2} \mathbf{I}\right) \xi^{(2)}=\mathbf{0}} \\ {\text { (b) Show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right) \xi^{(1)}=\left(r_{1}-r_{2}\right) \mathbf{\xi}^{(1)}} \\\ {\text { (c) Suppose that } \xi^{(1)} \text { and } \xi^{(2)} \text { are linearly dependent. Then } c_{1} \xi^{(1)}+c_{2} \xi^{(2)}=\mathbf{0} \text { and at least }}\end{array} $$ $$ \begin{array}{l}{\text { one of } c_{1} \text { and } c_{2} \text { is not zero; suppose that } c_{1} \neq 0 . \text { Show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2} \boldsymbol{\xi}^{(2)}\right)=\mathbf{0}} \\ {\text { and also show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2} \boldsymbol{\xi}^{(2)}\right)=c_{1}\left(r_{1}-r_{2}\right) \boldsymbol{\xi}^{(1)} \text { . Hence } c_{1}=0, \text { which is }} \\\ {\text { a contradiction. Therefore } \xi^{(1)} \text { and } \boldsymbol{\xi}^{(2)} \text { are linearly independent. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Modify the argument of part (c) in case } c_{1} \text { is zero but } c_{2} \text { is not. }} \\ {\text { (e) Carry out a similar argument for the case in which the order } n \text { is equal to } 3 \text { ; note that }} \\ {\text { the procedure can be extended to cover an arbitrary value of } n .}\end{array} $$

Short answer

**Question:** Show that the vectors \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly independent given the matrix equation \((\mathbf{A}-r_{i}\mathbf{I})\xi^{(i)} = \mathbf{0}\) for \(i=1,2\) and \(r_{1} \neq r_{2}\). **Answer:** We assumed that \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly dependent and showed that this assumption leads to contradictions in both cases when \(c_{1} \neq 0\) and when \(c_{1} = 0, c_{2} \neq 0\). Therefore, the vectors \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly independent.

Step-by-step solution

a) Matrix equation for \(\xi^{(1)}\) and \(\xi^{(2)}\)

Given the matrix equation \((\mathbf{A}-r_{i}\mathbf{I})\xi^{(i)} = \mathbf{0}\) for \(i=1,2\).

b) Show that \((\mathbf{A}-r_{2} \mathbf{I}) \xi^{(1)} = (r_{1}-r_{2})\mathbf{\xi}^{(1)}\)

First, multiply both sides of the given matrix equation for \(\xi^{(1)}\) by \(\mathbf{A}-r_{2}\mathbf{I}\): $$(\mathbf{A}-r_{2} \mathbf{I})(\mathbf{A}-r_{1} \mathbf{I})\xi^{(1)} = (\mathbf{A}-r_{2} \mathbf{I})\mathbf{0}$$ Now, we know that \(\xi^{(2)}\) satisfies the equation \((\mathbf{A}-r_{2}\mathbf{I})\xi^{(2)} = \mathbf{0}\), which means $$(\mathbf{A}-r_{1} \mathbf{I})\xi^{(1)} = \xi^{(2)}$$ So, substituting in the above equation, we get $$(\mathbf{A}-r_{2} \mathbf{I})\xi^{(1)} = (r_{1}-r_{2}) \mathbf{\xi}^{(1)}$$

c) Assume \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly dependent

Suppose that \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly dependent. Then, there exist scalars \(c_{1}\) and \(c_{2}\) such that $$c_{1} \xi^{(1)}+c_{2} \xi^{(2)} = \mathbf{0}$$ and at least one of \(c_{1}\) and \(c_{2}\) is nonzero. Assume \(c_{1} \neq 0\). Now, apply the matrix \((\mathbf{A}-r_{2} \mathbf{I})\) to both sides of the equation, $$(\mathbf{A}-r_{2} \mathbf{I})(c_{1} \xi^{(1)}+c_{2} \xi^{(2)}) = (\mathbf{A}-r_{2} \mathbf{I})\mathbf{0} = \mathbf{0}$$ From part (b), we have \((\mathbf{A}-r_{2} \mathbf{I}) \xi^{(1)} = (r_{1}-r_{2}) \mathbf{\xi}^{(1)}\), and since \((\mathbf{A}-r_{2} \mathbf{I})\xi^{(2)} = \mathbf{0}\), we can rewrite the above equation as $$c_{1}(r_{1}-r_{2})\mathbf{\xi}^{(1)} = \mathbf{0}$$ Since \(c_{1} \neq 0\) and \(r_{1} \neq r_{2}\), this implies that \(\xi^{(1)} = \mathbf{0}\), which is a contradiction. Therefore, \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly independent.

d) Modify the argument for the case \(c_{1} = 0\) and \(c_{2} \neq 0\)

If \(c_{1} = 0\) and \(c_{2} \neq 0\), then the equation \(c_{1} \xi^{(1)}+c_{2} \xi^{(2)} = \mathbf{0}\) becomes $$c_{2} \xi^{(2)} = \mathbf{0}$$ Now, apply the matrix \((\mathbf{A}-r_{1} \mathbf{I})\) to both sides of the equation, $$(\mathbf{A}-r_{1} \mathbf{I})(c_{2} \xi^{(2)}) = (\mathbf{A}-r_{1} \mathbf{I})\mathbf{0} = \mathbf{0}$$ Since \((\mathbf{A}-r_{1} \mathbf{I})\xi^{(1)} = \mathbf{0}\) and \((\mathbf{A}-r_{2} \mathbf{I}) \xi^{(2)} = (r_{1}-r_{2}) \mathbf{\xi}^{(2)}\), this equation becomes $$c_{2}(r_{1}-r_{2})\mathbf{\xi}^{(2)} = \mathbf{0}$$ Since \(c_{2} \neq 0\) and \(r_{1} \neq r_{2}\), this implies that \(\xi^{(2)} = \mathbf{0}\), which is a contradiction. Therefore, \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly independent.