Question

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}\end{aligned}} \right)\). Show that \({A^{\bf{3}}} = {\bf{0}}\). Use matrix algebra to complete the product \(\left( {I - A} \right)\left( {I + A + {A^{\bf{2}}}} \right)\).

Short answer

The equation \({A^3} = 0\) is proved. Using \({A^3} = 0\), the product \(\left( {I - A} \right)\left( {I + A + {A^2}} \right) = I\).

Step-by-step solution

First compute \({A^{\bf{2}}}\)

\(\begin{aligned}{c}{A^2} = AA\\ = \left( {\begin{aligned}{*{20}{c}}0&0&0\\1&0&0\\0&1&0\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}0&0&0\\1&0&0\\0&1&0\end{aligned}} \right)\\{A^2} = \left( {\begin{aligned}{*{20}{c}}0&0&0\\0&0&0\\1&0&0\end{aligned}} \right)\end{aligned}\)

Using \({A^{\bf{2}}}\) compute \({A^{\bf{3}}}\)

\(\begin{aligned}{c}{A^3} = {A^2}A\\ = \left( {\begin{aligned}{*{20}{c}}0&0&0\\0&0&0\\1&0&0\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}0&0&0\\1&0&0\\0&1&0\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{aligned}} \right)\\{A^3} = 0\end{aligned}\)

Use matrix algebra

\(\begin{aligned}{c}\left( {I - A} \right)\left( {I + A + {A^2}} \right) = {I^2} + IA + I{A^2} - AI - {A^2} - {A^3}\\ = I + A + {A^2} - A - {A^2} - {A^3}\\ = I - {A^3}\\ = I - 0\\\left( {I - A} \right)\left( {I + A + {A^2}} \right) = I\end{aligned}\)

\({A^3} = 0\) from step 2.