Homepage › Solution manuals › Sheldon Axler › Linear Algebra Done Right › Exercise 5.B.13

Exercise 5.B.13

Answers

Proof. By 5.21, $W$ is either ${0}$ or infinite-dimensional. Let $U$ be a subspace of $W$ invariant under $T$ . Then $T |_{U}$ also has no eigenvalues. But $T |_{U}$ is also an operator on a complex vector space, therefore $U$ is either ${0}$ or infinite-dimensional. □

celiopassos

2017-10-06 00:00

Exercise 5.B.13

Answers

Comments

Add answer