Homepage › Solution manuals › Sheldon Axler › Linear Algebra Done Right › Exercise 7.B.12

Exercise 7.B.12

Answers

Proof. Let $e_{1}, \dots, e_{n}$ be an orthonormal basis of $V$ consisting of eigenvectors of $T$ and let $λ_{1}, \dots, λ_{n}$ denote their corresponding eigenvalues. Choose an eigenvalue $λ^{'}$ of $T$ such that $| λ^{'} - λ |^{2}$ is minimized. There are $a_{1}, \dots, a_{n} \in 𝔽$ such that

v = a_{1} e_{1} + \dots + a_{n} e_{n} .

Thus, we have

\begin{aligned} 𝜖^{2} & > | | Tv - λv | |^{2} \\ = | ⟨ Tv - λv, e_{1} ⟩ |^{2} + \dots + | ⟨ Tv - λv, e_{n} ⟩ |^{2} \\ = | λ_{1} a_{1} - λ a_{1} |^{2} + \dots + | λ_{n} a_{n} - λ a_{n} |^{2} \\ = | a_{1} |^{2} | λ_{1} - λ |^{2} + \dots + | a_{n} |^{2} | λ_{n} - λ |^{2} \\ \geq | a_{1} |^{2} | λ^{'} - λ |^{2} + \dots + | a_{n} |^{2} | λ^{'} - λ |^{2} \\ = | λ^{'} - λ |^{2}, \end{aligned}

where the second and fifth lines follow from 6.30 (the fifth because $| | v | | = 1$ ). Taking the square root now yields the desired result. □

celiopassos

2017-10-06 00:00

Exercise 7.B.12

Answers

Comments

Add answer