Question Number 224305 by gregori last updated on 01/Sep/25
Answered by fkwow344 last updated on 01/Sep/25
$$\mathrm{Let}'\mathrm{s}\:\mathrm{set}\:\mathrm{as}\:\overset{\rightarrow} {\boldsymbol{\mathrm{v}}}_{\mathrm{1}} =\left(\mathrm{2},\mathrm{1}\right)^{\intercal} \:,\:\overset{\rightarrow} {\boldsymbol{\mathrm{v}}}_{\mathrm{2}} =\left(\mathrm{1},\mathrm{0}\right)^{\intercal} \\ $$$${A}={PJP}^{−\mathrm{1}} \:\left(\mathrm{Jordan}\:\mathrm{decomposition}\right) \\ $$$${P}=\begin{pmatrix}{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{pmatrix}\:,\:{J}=\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}\:,\:{P}^{−\mathrm{1}} =\begin{pmatrix}{\mathrm{0}}&{\:\:\:\:\mathrm{1}}\\{\mathrm{1}}&{−\mathrm{2}}\end{pmatrix} \\ $$$$\mathrm{and}\:{J}^{{n}} =\begin{pmatrix}{\mathrm{1}}&{{n}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{So},\:{A}^{{n}} ={PJ}^{{n}} {P}^{−\mathrm{1}} \\ $$$$\therefore{A}^{{n}} =\begin{pmatrix}{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}}&{{n}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{0}}&{\:\:\:\:\mathrm{1}}\\{\mathrm{1}}&{−\mathrm{2}}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}&{−\mathrm{4}{n}}\\{\:\:\:\:\:\:{n}}&{\mathrm{1}−\mathrm{2}{n}}\end{pmatrix} \\ $$