Exercise 7.B.3

Answers

Proof. Define $T \in L (ℂ^{3})$ by

\begin{aligned} T e_{1} & = 2 e_{1} \\ T e_{2} & = 3 e_{2} \\ T e_{3} & = 3 (e_{1} + e_{2} + e_{3}), \end{aligned}

where $e_{1}, e_{2}, e_{3}$ is the standard basis of $ℂ^{3}$ . The matrix of $T$ with respect to same basis is

M (T) = (\begin{matrix} 2 & 0 & 3 \\ 0 & 3 & 3 \\ 0 & 0 & 3 \end{matrix}) .

Therefore, the only eigenvalues of $T$ are $2$ and $3$ (by 5.32). Moreover

\begin{aligned} (T^{2} - 5 T + 6 I) e_{3} & = T^{2} e_{3} - 5 T e_{3} + 6 I e_{3} \\ = T (3 e_{1} + 3 e_{2} + 3 e_{3}) - 15 (e_{1} + e_{2} + e_{3}) + 6 e_{3} \\ = 6 e_{1} + 9 e_{2} + 9 (e_{1} + e_{2} + e_{3}) - 15 (e_{1} + e_{2} + e_{3}) + 6 e_{3} \\ = 3 e_{2} \\ \neq 0 . \end{aligned}

Therefore, $T^{2} - 5 T + 6 I \neq 0$ . □

celiopassos

2017-10-06 00:00