Exercise 2.4.21

Answers

We have

A^{2} = [\begin{matrix} 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}], A^{3} = [\begin{matrix} 0 & 0 & 0 & 8 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}],

and $A^{4} =$ zero matrix for strictly triangular $A$ . Then

A v = A [\begin{matrix} x \\ y \\ z \\ t \end{matrix}] = [\begin{matrix} 2 y \\ 2 z \\ 2 t \\ 0 \end{matrix}], A^{2} v = [\begin{matrix} 4 z \\ 4 t \\ 0 \\ 0 \end{matrix}], A^{3} v = [\begin{matrix} 8 t \\ 0 \\ 0 \\ 0 \end{matrix}], A^{4} v = 0 .

mathchakra

2021-12-06 17:42