Question: Let \({\rm{u}}\) and \({\rm{v}}\) be the eigenvectors of a matrix \(A\), with corresponding eigenvalues \(\lambda \) and \(\mu \), and let \({c_1}\) and \({c_2}\) be scalars. Define \({{\rm{x}}_k} = {c_1}{\lambda ^k}{\rm{u}} + {c_2}{\mu ^k}{\rm{v}}\,\,\,\,\,\,\left( {k = 0,1,2,...} \right)\).

1. What is \({{\rm{x}}_{k + 1}}\), by definition?
2. Compute \({\rm{A}}{{\rm{x}}_k}\) from the formula for \({{\rm{x}}_k}\), and show that \(A{{\rm{x}}_k} = {{\rm{x}}_{k + 1}}\). This calculation will prove that the sequence \(\left\{ {{{\rm{x}}_k}} \right\}\) defined above satisfies the difference equation \({{\rm{x}}_{k + 1}} = A{{\rm{x}}_k}\,\,\,\,\left( {k = 0,1,2,...} \right)\).

(a)

Substitute \(k = k + 1\) in the formula of \({x_k} = {c_1}{\lambda ^k}u + {c_2}{\mu ^k}v\) as follows:

\({x_{k + 1}} = {c_1}{\lambda ^{k + 1}}u + {c_2}{\mu ^{k + 1}}v\)

Thus, \({x_{k + 1}} = {c_1}{\lambda ^{k + 1}}u + {c_2}{\mu ^{k + 1}}v\).

(b)

Substitute \({x_k} = {c_1}{\lambda ^k}u + {c_2}{\mu ^k}v\) in \(A{x_k}\) and solve as follows:

\(\begin{gathered} A{x_k} = A\left( {{c_1}{\lambda ^k}u + {c_2}{\mu ^k}v} \right) \\ = {c_1}{\lambda ^k}Au + {c_2}{\mu ^k}Av\,\,\,{\text{(}}\because {\text{by linearity)}} \\ = {c_1}{\lambda ^k}\lambda u + {c_2}{\mu ^k}\mu v\,\,{\text{(}}\because u{\text{ and }}v{\text{ are eigenvectors)}} \\ = {c_1}{\lambda ^{k + 1}}u + {c_2}{\mu ^{k + 1}}v \\ = {x_{k + 1}} \\ \end{gathered} \)

Hence, \(A{x_k} = {x_{k + 1}}\).