Exercise 5.A.17

Answers

Proof. Let $v_{1}, \dots, v_{4}$ be a basis of $ℝ^{4}$ . Define $T$ by

\begin{aligned} T v_{1} & = i v_{1} \\ T v_{2} & = (1 + i) v_{2} \\ T v_{3} & = (π - i) v_{3} \\ T v_{4} & = (\sqrt{2} + i) v_{4} . \end{aligned}

Hence $T$ has $4$ non-real eigenvalues. Since $\dim ℝ^{4} = 4$ , $T$ has no other eigenvalues. □

celiopassos

2017-10-06 00:00