Exercise 7.B.10

Answers

Proof. Define $T \in L (ℝ^{2})$ by

\begin{aligned} T e_{1} & = e_{2} \\ T e_{2} & = - e_{1}, \end{aligned}

where $e_{1}, e_{2}$ is the standard basis of $ℝ^{2}$ . Then

(T^{2} + I) e_{1} = T^{2} e_{1} + I e_{1} = T e_{2} + e_{1} = - e_{1} + e_{1} = 0 .

Therefore $T^{2} + bT + cI$ is not injective for $b = 0$ and $c = 1$ . □

celiopassos

2017-10-06 00:00