Exercise 8.C.5

Answers

Proof. Let

A = (\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 3 \end{matrix}) .

Then

\begin{aligned} {(A - I)}^{2} (A - 3) A & = {(\begin{matrix} - 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 \end{matrix})}^{2} (\begin{matrix} - 3 & 0 & 0 & 0 \\ 0 & - 2 & 1 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) (\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 3 \end{matrix}) \\ = {(\begin{matrix} - 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 \end{matrix})}^{2} (\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & - 2 & - 1 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) \\ = (\begin{matrix} - 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 \end{matrix}) (\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) \\ = (\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) . \end{aligned}

Define $T \in L (ℂ^{4})$ by

T (z_{1}, z_{2}, z_{3}, z_{4}) = (0, z_{2} + z_{3}, z_{3}, 3 z_{4}) .

Then $M (T) = A$ . The rest of this exercise is almost the same as the previous one. □

celiopassos

2017-10-06 00:00