Homepage › Solution manuals › Sheldon Axler › Linear Algebra Done Right › Exercise 5.B.2

Exercise 5.B.2

Answers

Proof. Suppose $λ$ is an eigenvalue of $T$ and $v$ a corresponding eigenvector. Then

\begin{aligned} 0 & = (T - 2 I) (T - 3 I) (T - 4 I) v \\ = (T^{3} - 9 T^{2} + 26 T - 24 I) v \\ = T^{3} v - 9 T^{2} v + 26 T - 24 v \\ = λ^{3} v - 9 λ^{2} v + 26 λv - 24 v \\ = (λ^{3} - 9 λ^{2} + 26 λ - 24) v \end{aligned}

Therefore, because $v \neq 0$ , $λ^{3} - 9 λ^{2} + 26 λ + 24 = 0$ . But we can factor this to $(λ - 2) (λ - 3) (λ - 4) = 0$ . Thus $λ$ is either $2$ , $3$ or $4$ . □

celiopassos

2017-10-06 00:00

Exercise 5.B.2

Answers

Comments

Add answer